Show that $x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n$ is a bounded sequence.

Jakobian gave a very good hint. I am giving somewhat sharper bounds for $x_n$. Note that the OP incorrectly used the Cauchy–Bunyakovsky–Schwarz inequality. Indeed, when applied correctly, we get $$\sqrt{n}\left(x_n+\sqrt{n}\right)=\sqrt{\sum_{k=1}^n1^2}\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2}\geq \sum_{k=1}^n1\cdot\left(\frac{k+1}{k}\right)=n+H_n,$$ where $H_n=\sum_{k=1}^n\frac1k$ is the $n$th harmonic number. That is, $$x_n\geq \frac{H_n}{\sqrt{n}}.$$

For the upper bound, the OP got $$x_n=\frac{\sum_{k=1}^n \frac{1}{k^2}+2H_n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2}+\sqrt{n}}.$$ We can show that $\sum_{k=1}^n\frac{1}{k^2}$ is bounded above by $2$ via $$\sum_{k=1}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\left(1-\frac{1}{n}\right)<2.$$ However, a sharper upper bound is $\sum_{k=1}^n\frac{1}{k^2}<\zeta(2)=\frac{\pi^2}{6}$. Also, $$\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2\geq \sum_{k=1}^n1^2=n.$$ This proves that $$x_n<\frac{\frac{\pi^2}{6}+2H_n}{\sqrt{n}+\sqrt{n}}=\frac{H_n}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right).$$ (You can replace $\frac{\pi^2}{6}$ by $2$ if you want a precalculus solution.)

That is, we have $$\frac{H_n}{\sqrt{n}}\leq x_n<\frac{H_n}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right).$$ So, $x_n$ has the asymptotic behavior of $\frac{H_n}{\sqrt{n}}\approx \frac{\ln n}{\sqrt{n}}$. But to show that $x_n$ is bounded, we don't need to know that $H_n\approx \ln n$ (well, to be even more precise, $H_n\approx \gamma+\ln n$, where $\gamma$ is the Euler–Mascheroni constant). Clearly, $$\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right)\leq \frac{\pi^2}{6}\left(\frac{1}{2}\right).$$ Furthermore, $$H_n=\sum_{k=1}^n\frac{1}{k}\leq \sum_{k=1}^n\frac{1}{\sqrt{k}}<\sum_{k=1}^n\frac{2}{\sqrt{k}+\sqrt{k-1}}=2\sum_{k=1}^n(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}.$$ Hence, $$x_n<\frac{2\sqrt{n}}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2}\right)=2+\frac{\pi^2}{6}\left(\frac{1}{2}\right)<\infty.$$ Again, replace $\frac{\pi^2}{6}$ by $2$ if you don't want any non-precalculus knowledge in the proof. So, you would get $x_n<3$ for all $n$.

Even if $\sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2 = n + 2H_n + H_n^{(2)}$ involves (generalized) harmonic numbers, it is pretty straightforward to notice$^{(*)}$ that $H_n=O(\log n)$ and $H_{n}^{(2)}=O(1)$ as $n\to +\infty$, hence

$$ \sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2 = n\left(1+O\left(\frac{\log n}{n}\right)\right),\tag{1}$$

$$ \sqrt{\sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2} = \sqrt{n}\left(1+O\left(\frac{\log n}{n}\right)\right)=\sqrt{n}+O\left(\frac{\log n}{\sqrt{n}}\right)\tag{2}$$ and the wanted limit is zero.

$(*)$ $H_n \leq 1+\int_{1}^{n}\frac{dx}{x}=1+\log(n)$ and $H_{n}^{(2)}\leq 1+\sum_{k=1}^{n}\frac{1}{k(k+1)}\leq 2$.