Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$
Here's a derivation that, while fairly long, is self-contained and uses only basic series manipulation techniques, like partial fractions decomposition, telescoping, swapping the order of summation, etc. It leans heavily on ideas from Borwein and Girgensohn's paper "Evaluation of Triple Euler Sums" (Electronic Journal of Combinatorics 3(1) 1996).
First, some notation. Define the multiple zeta functions by \begin{align} \zeta_N(a) &= \sum_{x=1}^N \frac{1}{x^a}, \:\:\: \zeta_N(a,b) = \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^a y^b}, \:\:\: \zeta_N(a,b,c) = \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c}, \\ \zeta(a,b) &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^a y^b}, \:\:\: \zeta(a,b,c) = \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c}. \end{align}
We will need the following symmetry relation, as well as expressions for $\zeta(4,1)$ and $\zeta(2,2,1) + \zeta(2,1,2)$. Proofs for all of these are given at the end of the post. \begin{align} \zeta_N(a,b) + \zeta_N(b,a) &= \zeta_N(a) \zeta_N(b) - \zeta_N(a+b) \tag{1}\\ \zeta(4,1) &= \zeta(5) - \zeta(3,2) - \zeta(2,3) \tag{2}\\ \zeta(2,2,1) + \zeta(2,1,2) &= \zeta(2,3) + \zeta(3,2) \tag{3} \end{align}
Given these, we have
The Main Proof: \begin{align} \sum_{k=1}^{\infty} \frac{H^{(2)}_k H_k}{k^2} &= \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} + \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1}}{k^3} + \sum_{k=1}^{\infty} \frac{H_{k-1}}{k^4} + \sum_{k=1}^{\infty} \frac{1}{k^5} \\ &= \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} + \zeta(3,2) + \zeta(4,1) + \zeta(5). \end{align} The most complicated sum is the first, so let's look at that more closely. \begin{align} \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} &= \sum_{k=1}^{\infty} \frac{1}{k^2} \zeta_{k-1}(2) \zeta_{k-1}(1) \\ &= \sum_{k=1}^{\infty} \frac{1}{k^2} (\zeta_{k-1}(2,1) + \zeta_{k-1}(1,2) + \zeta_{k-1}(3)), \text{ by (1)} \\ &= \zeta(2,2,1) + \zeta(2,1,2) + \zeta(2,3), \text{ by definition of the multiple zeta functions} \\ &= 2\zeta(2,3) + \zeta(3,2), \text{ by (3)}. \end{align} Thus \begin{align} \sum_{k=1}^{\infty} \frac{H^{(2)}_k H_k}{k^2} &= 2 \zeta(2,3) + \zeta(3,2) + \zeta(3,2) + \zeta(5) - \zeta(3,2) - \zeta(2,3) + \zeta(5), \text{ by (2)} \\ &= \zeta(2,3) + \zeta(3,2) + 2 \zeta(5) \\ &= \zeta(2) \zeta(3) - \zeta(5) + 2 \zeta(5), \text{ by (1)} \\ &= \zeta(2) \zeta(3) + \zeta(5). \end{align}
Proof of (1): \begin{align} \zeta_N(a,b) + \zeta_N(b,a) &= \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^a y^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a} \\ &= \sum_{y=1}^N \sum_{x=y+1}^N \frac{1}{x^a y^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}, \\ & \:\:\:\:\:\: \text{ swapping the order of summation on the first sum} \\ &= \sum_{x=1}^N \sum_{y=x+1}^N \frac{1}{y^a x^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}, \text{ relabeling variables on the first sum} \\ &= \sum_{x=1}^N \sum_{y=1}^N \frac{1}{y^a x^b} - \sum_{x=1}^N \frac{1}{x^{a+b}}, \text{ combining sums} \\ &= \zeta_N(a) \zeta_N(b) - \zeta_N(a+b). \square \end{align}
Proof of (2): \begin{align} \zeta(4,1) &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^4 y} \\ &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^4 (x-y)}, \text{ reindexing the second sum} \\ &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \left(-\frac{1}{x^4 y} - \frac{1}{x^3 y^2} - \frac{1}{x^2y^3} - \frac{1}{x y^4} + \frac{1}{(x-y)y^4}\right), \\ &\:\:\:\:\: \text{ by partial fractions decomposition}\\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \left(\frac{1}{(x-y)y^4} - \frac{1}{x y^4} \right) \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{y^4} \left(\frac{1}{x-y} - \frac{1}{x} \right) \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{y=1}^{\infty} \frac{1}{y^4} \sum_{x=y+1}^{\infty} \left(\frac{1}{x-y} - \frac{1}{x} \right), \\ & \:\:\:\:\: \text{ swapping the order of summation} \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{y=1}^{\infty} \frac{1}{y^4} \sum_{x=1}^y \frac{1}{x}, \text{ as the sum telescopes} \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \zeta(4,1) + \zeta(5) \\ &= \zeta(5) - \zeta(3,2) - \zeta(2,3). \square \end{align}
For the proof of (3), we need the following additional symmetry result: \begin{equation} \zeta_N(a,b,c) + \zeta_N(a,c,b) + \zeta_N(c,a,b) = \zeta_N(c) \zeta_N(a,b) - \zeta_N(a,b+c) - \zeta_N(a+c,b) \tag{4} \end{equation}
Proof of (4): \begin{align} &\zeta_N(a,b,c) + \zeta_N(a,c,b) + \zeta_N(c,a,b) \\ &=\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b} \\ &= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=z+1}^{x-1} \frac{1}{x^a y^b z^c} + \sum_{y=1}^N \sum_{x=y+1}^N \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\ &\:\:\:\:\:\text{ swapping order of summation on the first two sums} \\ &= \sum_{z=1}^N \sum_{x=z+1}^N \sum_{y=z+1}^{x-1} \frac{1}{x^a y^b z^c} + \sum_{y=1}^N \sum_{x=y+1}^N \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\ &\:\:\:\:\:\text{ swapping order of summation on the first sum} \\ &= \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=x+1}^{y-1} \frac{1}{x^c y^a z^b} + \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=1}^{x-1}\frac{1}{x^c y^a z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\ &\:\:\:\:\: \text{ relabeling variables on the first two sums} \\ &= \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=1}^{y-1} \frac{1}{x^c y^a z^b} - \sum_{x=1}^N \sum_{y=x+1}^N \frac{1}{x^{b+c} y^a} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\ &\:\:\:\:\: \text{ combining the first two sums} \\ &= \sum_{x=1}^N \sum_{y=1}^N \sum_{z=1}^{y-1} \frac{1}{x^c y^a z^b} - \sum_{x=1}^N \sum_{z=1}^{y-1} \frac{1}{x^{a+c} z^b} - \sum_{y=1}^N \sum_{x=1}^{y-1} \frac{1}{x^{b+c} y^a}, \\ &\:\:\:\:\:\text{ combining the first and third sums and swapping the order of summation on the second} \\ &= \zeta_N(c) \zeta_N(a,b) - \zeta_N(a+c,b) - \zeta_N(a,b+c). \square \end{align}
Proof of (3): \begin{align} \zeta_N(2,2,1) &= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2 y^2 z} \\ &= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2 y^2 (y-z)}, \text{ reindexing on the third sum} \\ &= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \left( -\frac{1}{x^2 y z^2} - \frac{1}{x^2 y^2 z} + \frac{1}{x^2(y-z)z^2} \right), \\ &\:\:\:\:\: \text{ by partial fractions decomposition} \\ &= - \zeta_N(2,1,2) - \zeta_N(2,2,1) + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2(y-z)z^2} \tag{5}. \\ \end{align} Now, let's look at the third expression in (5). \begin{align} &\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2(y-z)z^2} \\ &= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{x-y-1} \frac{1}{x^2(x-y-z)z^2}, \text{ reindexing the second sum} \\ &= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=y+1}^{x-1} \frac{1}{x^2(x-z)(z-y)^2}, \text{ reindexing the third sum} \\ &= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \frac{1}{x^2(x-z)(z-y)^2}, \text{ swapping the order of summation} \\ &= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \frac{1}{x^2(x-z)y^2}, \text{ reindexing the third sum} \\ &= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \left(-\frac{1}{x y^2 z^2} - \frac{1}{x^2 y^2 z} + \frac{1}{(x-z)y^2 z^2} \right), \text{ by partial fractions decomposition} \\ &= - \zeta_N(1,2,2) - \zeta_N(2,1,2) + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} \tag{6}, \text{ relabeling variables}. \end{align} Let's look at the third expression in (6). \begin{align} &\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} \\ &= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} + \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} - \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} \\ &= \left(\sum_{x=1}^N \frac{1}{x}\right) \left(\sum_{y=1}^N \sum_{z=1}^{y-1} \frac{1}{y^2 z^2} \right) - \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2}, \\ &\:\:\:\:\: \text{ via the finite sum version of the Cauchy product} \\ &= \zeta_N(1) \zeta_N(2,2) - e_N(1,2,2), \tag{7} \\ \end{align} where $$e_N(1,2,2) = \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2}.$$ Putting (5), (6), and (7) together, we have \begin{align} \zeta_N(2,2,1) =& - \zeta_N(2,1,2) - \zeta_N(2,2,1) - \zeta_N(1,2,2) - \zeta_N(2,1,2) + \zeta_N(1) \zeta_N(2,2) \\ &- e_N(1,2,2), \\ \zeta_N(2,2,1) + \zeta_N(2,1,2) &= - \zeta_N(1) \zeta_N(2,2) + \zeta_N(2,3) + \zeta_N(3,2) + \zeta_N(1) \zeta_N(2,2) \\ &- e_N(1,2,2), \text{ by (4)} \\ =& \zeta_N(2,3) + \zeta_N(3,2) - e_N(1,2,2). \\ \end{align} All that remains to complete the proof of (3) is to show that $e_N(1,2,2) \to 0$ as $N \to \infty$. We have \begin{align} e_N(1,2,2) &= \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} \\ &\leq \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^N \frac{1}{x y^2 z^2} \\ &= \zeta_N(2) \sum_{x=1}^N \sum_{y=N+1-x}^N \frac{1}{x y^2} \\ &= \zeta_N(2) \sum_{y=1}^N \sum_{x=N+1-y}^N \frac{1}{x y^2}, \text{ swapping the order of summation} \\ &\leq \zeta_N(2) \sum_{y=1}^N \frac{1}{y^2} \sum_{x=N+1-y}^N \frac{1}{N+1-y} \\ &= \zeta_N(2) \sum_{y=1}^N \frac{1}{y^2} \frac{y}{N+1-y} \\ &= \zeta_N(2) \sum_{y=1}^N \frac{1}{y (N+1-y)}\\ &= \zeta_N(2) \frac{1}{N+1}\sum_{y=1}^N \left(\frac{1}{y} + \frac{1}{N+1-y} \right), \text{ by partial fractions decomposition} \\ &= \zeta_N(2) \frac{2}{N+1} \zeta_N(1), \end{align} which goes to $0$ as $N \to \infty$, since $\zeta_N(1) = O(\log N)$ and $\zeta_N(2) = O(1)$. $\square$
Consider the integral $$I= - \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} (\sin^{-1} x)^4 \,dx.$$
Since $$(\sin^{-1} x)^4 = \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2 n} \tag{1}$$ and $$-\int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx= \frac{\pi}{2} \binom{2n}{n} \frac{(H_n + 2\ln2)}{2^{2n}}, \tag{2}$$
we have $$\begin{align*} &I= - \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx \\&= \frac{3 \pi}{4} \sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}}{n^2} ( H_n +2 \ln2 ) \\& = \frac{\pi^5}{80} \ln2 + \frac{3 \pi}{4} \sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_n}{n^2}. \end{align*}$$
However, substituting $x\mapsto \sin x$ and employing the fourier expansion of $\ln \cos x$: $$\begin{align*} & I= -2 \int_0^{\pi/2} x^4 \, \ln\cos x\, dx \\&= 2 \int_0^{\pi/2} x^4 \left(\ln2 + \sum_{n=1}^{\infty} \frac{(-1)^n \cos(2 x n)}{n} \right)dx \\&= \frac{\pi^5}{80}\ln2 + 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \int_0^{\pi/2} x^4 \cos(2 x n) dx \\&= \frac{\pi^5}{80}\ln2 + 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \frac{(-1)^n}{n^2}\left(\frac{\pi^3}{8}-\frac{3 \pi}{4 n^2}\right) \\&= \frac{\pi^5}{80}\ln2 + \frac{\pi^3}{4}\zeta(3) - \frac{3 \pi}{2} \zeta(5). \end{align*}$$
Therefore,
$$\sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_n}{n^2} = 2\zeta(2)\,\zeta(3)-2\zeta(5).$$
Finish off using Euler's formula for $\sum H_n/n^q $.
Notes.
You may find a proof of $(1)$ here, and $(2)$ is just the derivative of a beta function. The swap of the sum and the integral should be justified.
I found this proof while exploring series involving $H_n^{(2)}$. Using the same method, I also obtained the following related results: $$\sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_{2n}}{n^2} =\frac{11}{4}\zeta(2)\,\zeta(3)-\frac{47}{16}\zeta(5) \tag{3}$$ $$\sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_{n}^2}{n^2} = 4 \zeta(3)^2 - \frac{5}{8} \zeta(6) \tag{4}$$ $$\sum_{n=1}^{\infty} \frac{H_n \left(H_{n-1}^{(2)2}-H_{n-1}^{(4)}\right)}{n^2} = 3\,\zeta(3)\,\zeta(4)-4\,\zeta(2)\,\zeta(5)+4\,\zeta(7) \tag{5}$$ and others.
I think it is reasonable to start with: $$\sum_{k=1}^{+\infty}\frac{H_k^{(2)}H_k}{k^2}=\sum_{k=1}^{+\infty}\frac{H_k}{k^4}+\sum_{k=1}^{+\infty}\frac{H_k}{k^2}\sum_{1\leq j< k}\frac{1}{j^2},\tag{1}$$ that leads to: $$\sum_{k=1}^{+\infty}\frac{H_k^{(2)}H_k}{k^2}=\left(\sum_{k=1}^{+\infty}\frac{H_k}{k^2}\right)\left(\sum_{j=1}^{+\infty}\frac{1}{j^2}\right)-\sum_{k=1}^{+\infty}\frac{1}{k^2}\sum_{1\leq j< k}\frac{H_j}{j^2},\tag{2}$$ Now since: $$\operatorname{Li}_2(x)+\frac{\log^2(1-x)}{2}=\sum_{k=1}^{+\infty}\frac{H_k}{k}x^k,\tag{3}$$ $$\frac{\log^2(1-x)}{2}=\sum_{k=1}^{+\infty}\frac{H_{k-1}}{k}x^k,\tag{4}$$ follows. By dividing by $x$ and integrating between $0$ and $1$ we get: $$\sum_{k=1}^{+\infty}\frac{H_{k-1}}{k^2}=\frac{1}{2}\int_{0}^{1}\frac{\log^2(x)}{1-x}dx=\frac{1}{2}\int_{0}^{+\infty}\frac{u^2}{e^u-1}du=\zeta(3),\tag{5}$$ so: $$\sum_{k=1}^{+\infty}\frac{H_k^{(2)}H_k}{k^2}=2\zeta(2)\zeta(3)-\sum_{k=1}^{+\infty}\frac{1}{k^2}\sum_{1\leq j< k}\frac{H_j}{j^2}.\tag{6}$$ For the last term consider: $$-\frac{\log(1-xy)}{y(1-xy)}=\sum_{k=1}^{+\infty}H_k x^k y^{k-1}, \tag{7}$$ multiply both terms by $-\log(y)$ and integrate between $0$ and $1$ with respect to $y$: $$\int_{0}^{1}\frac{\log(y)\log(1-xy)}{y(1-xy)}dy = \sum_{k=1}^{+\infty}\frac{H_k}{k^2}x^k.\tag{8}$$ Multiplying both sides by $-\frac{\log x}{1-x}$ and integrating between $0$ and $1$ with respect to $x$ should do the trick. For the last part it is only required to find an appropriate birational diffeomorphism of the unity square that puts the integral in a nicer form - a sort of "reverse Viola-Rhin method".