Show $(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots)^2 = 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49} + \cdots$

Let's have a try. $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{4n+1}=\int_{0}^{1}\frac{dx}{1+x^4},\qquad S=\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)=\int_{0}^{1}\frac{1+x^2}{1+x^4}dx,$$ $$ S = \int_{0}^{1}\frac{x+x^{-1}}{x^{-2}+x^2}\frac{dx}{x}=\int_{1}^{+\infty}\frac{z}{(2z^2-1)\sqrt{1-z^2}}\,dz = \int_{0}^{1}\frac{dt}{(2-t^2)\sqrt{1-t^2}},$$ $$ S = \int_{0}^{\pi/2}\frac{d\theta}{2-\sin^2\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta}=\frac{1}{2}\int_{\mathbb{R}}\frac{du}{2+u^2},$$ where in the last integral we used the substitution $\theta=\arctan u$. This gives: $$ S^2 = \frac{1}{8}\int_{\mathbb{R}^2}\frac{du\,dv}{(1+u^2)(1+v^2)}=\int_{0}^{1}\int_{0}^{+\infty}\frac{1}{(1+z^2)(1+x^2)}dx\,dz$$ On the other hand, $$\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\int_{0}^{1}\frac{\log y}{y^2-1}dy=\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2y^2)}dx\,dy,$$ where I learned the last equality from the Mike Spivey's note on the Luigi Pace's proof of $\zeta(2)=\frac{\pi^2}{6}$, just here. By setting $y=\frac{z}{x}$ in the last integral we get $S^2=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}$, QED. So it looks like @user17762's proof-by-squaring-the-arctangent-series and Pace's proof can be combined in order to get a very short proof of your claim.

For the sake of exposing a one-line-proof of $\zeta(2)=\frac{\pi^2}{6}$: $$\zeta(2)=\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy=\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\frac{\pi^2}{6}.$$