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
$$ I = \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx. $$
On one hand, we observe
$$ \frac{1}{(1+x^2)(1+y^2)} = \sum_{n \geq 0} \left(-x^2 \right)^n \sum_{m \geq 0} \left(-y^2 \right)^m, \quad |x|,|y| < 1. $$
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
$$ I = \int_{1}^{\infty}\int_{1}^{\infty} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx, $$
This implies
$$ I = \frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx. $$
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
$$ I = \frac{1}{4}\int_{0}^{\infty} \frac{\log(t)}{t^2-1} \ dt. $$
Now, observe
$$ \int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt + \int_{1}^{\infty} \frac{\log(t)}{t^2-1} \ dt= \int_{0}^{\infty} \frac{\log(t)}{t^2-1} \ dt. $$
We see that
$$ \int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt = \int_{1}^{\infty} \frac{\log(t)}{t^2-1} \ dt $$
by making a change of variables $ t \mapsto 1/t $ on the second integral. Thus,
$$ I = \frac{1}{2} \int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt. $$
Now, convert the integrand into a geometric series
$$ \frac{\log(t)}{t^2-1} = - \sum_{n \geq 0} \log(t) t^{2n}, \quad t \in (0,1), $$
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
$$ I = \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}. $$
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. See https://www.ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3/