Here is a proof of the identity
$$\left(\sum_{n \geq 0} \frac{(-1)^n}{(2n+1)}\right)^2 = \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}$$
without knowing the value of either series.
Consider the double integral
\begin{align*}
I &= \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx.
\end{align*}
On one hand, we observe
\begin{align*}
\frac{1}{(1+x^2)(1+y^2)} &= \sum_{n \geq 0} \left(-x^2 \right)^n \sum_{m \geq 0} \left(-y^2 \right)^m, \quad |x|,|y| < 1.
\end{align*}
Then, we get that
\begin{align*}
I &= \int_{0}^1 \int_{0}^1 \sum_{n \geq 0} \left(-x^2 \right)^n \sum_{m \geq 0} \left(-y^2 \right)^m \ dy \ dx \\
&=\sum_{n \geq 0}\sum_{m \geq 0} \int_{0}^1 \int_{0}^1 \left(-x^2 \right)^n \left(-y^2 \right)^m \ dy \ dx \\
&=\sum_{n \geq 0}\sum_{m \geq 0} \frac{(-1)^n}{2n+1} \frac{(-1)^m}{2m+1} \\
&=\left(\sum_{n \geq 0} \frac{(-1)^n}{2n+1} \right)^2.
\end{align*}
The interchanging of sum and integral done here can be justified by the Dominated Convergence Theorem.
Now, we make the substitution $x \mapsto 1/x,$ and $y \mapsto 1/y$ to $I,$ resulting in
\begin{align*}
I &= \int_{1}^{\infty}\int_{1}^{\infty} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx,
\end{align*}
This implies
\begin{align*}
I &= \frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx.
\end{align*}
Make another substitution $y=tx$ with $dy = x \ dt$ to see
\begin{align*}
I &= \frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{x}{(1+x^2)(1+t^2x^2)} \ dt \ dx \\
&=\frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{x}{(1+x^2)(1+t^2x^2)} \ dx \ dt,
\end{align*}
in which the interchanging of the integral is justified by Tonelli's Theorem.
Now, using the partial fractions identity,
$$\frac{x}{(1+x^2)(1+t^2x^2)} = \frac{2 x}{\left(2-2 t^2\right) \left(x^2+1\right)}-\frac{2 t^2 x}{\left(2-2 t^2\right)
\left(t^2 x^2+1\right)},$$
and integrating with respect to $x,$ we get
\begin{align*}
I &= \frac{1}{4}\int_{0}^{\infty} \frac{\log(t)}{t^2-1} \ dt.
\end{align*}
Now, observe
\begin{align*}
\int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt + \int_{1}^{\infty} \frac{\log(t)}{t^2-1} \ dt= \int_{0}^{\infty} \frac{\log(t)}{t^2-1} \ dt.
\end{align*}
We see that
\begin{align*}
\int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt &= \int_{1}^{\infty} \frac{\log(t)}{t^2-1} \ dt
\end{align*}
by making a change of variables $t \mapsto 1/t$ on the second integral. Thus,
\begin{align*}
I &= \frac{1}{2} \int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt.
\end{align*}
Now, convert the integrand into a geometric series
\begin{align*}
\frac{\log(t)}{t^2-1} &= - \sum_{n \geq 0} \log(t) t^{2n}, \quad t \in (0,1),
\end{align*}
and we see that
\begin{align*}
I &= -\frac{1}{2} \int_{0}^{1} \sum_{n \geq 0} \log(t) t^{2n} \ dt \\
&= -\frac{1}{2} \sum_{n \geq 0} \int_{0}^{1} \log(t) t^{2n} \ dt \\
&= \frac{1}{2} \sum_{n \geq 0} \frac{1}{(2n+1)^2},
\end{align*}
in which the interchanging of sum and integral is justified by Beppo Levi's Monotone Convergence Theorem.
Observing that
\begin{align*}
\sum_{n \in \mathbb{N}} \frac{1}{n^2} &= \sum_{n \in \mathbb{N}} \frac{1}{(2n)^2}+\sum_{n \geq 0} \frac{1}{(2n+1)^2} \\
&=
\frac{1}{4}\sum_{n \in \mathbb{N}} \frac{1}{n^2}+\sum_{n \geq 0} \frac{1}{(2n+1)^2},
\end{align*}
we find
\begin{align*}
I &= \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}.
\end{align*}
Hence, we have
$$\left(\sum_{n \geq 0} \frac{(-1)^n}{(2n+1)}\right)^2 = \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}.$$
Remark
For those that may be interested, these types of integral calculations to evaluate series may be found in my joint paper with Daniele Ritelli, "Evaluation of harmonic sums with integrals", Q. Appl. Math. 76, No. 3, 577-600 (2018), MR3805043, Zbl 1397.40001.
