Adding to the last paragraph of Raymond Manzoni's answer. The given equality can be derived from the finite version $ 2\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n^{2}}+\sum_{k=1}^{N}\frac{(-1)^{N+k-1}}{ k^{2}\dbinom{N}{k}\dbinom{N+k}{k}}=3\sum_{n=1}^{N}\frac{1}{n^{2}\dbinom{2n}{n}}.\tag{1} $
In the footnote 4 of Alf van der Poorten's article A proof that Euler missed ... Apéry's proof of the irrationality of $\zeta (3)$, the author states that the following identity $ \zeta (2):=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6} =3\sum_{n=1}^{\infty }\frac{1}{n^{2}\dbinom{2n}{n}}\tag{2} $ may be proved by slightly varying the argument in section 3 - multiply by $ (-1)^{n-1}$ instead of dividing by $n$. In this section 3 the equivalent one for $\zeta (3)$ is proved $ \zeta (3):=\sum_{n=1}^{\infty }\frac{1}{n^{3}}=\frac{5}{2}\sum_{n=1}^{\infty } \frac{(-1)^{n-1}}{n^{3}\dbinom{2n}{n}},\tag{3} $ as a consequence of$^1$ $ \sum_{n=1}^{N}\frac{1}{n^{3}}=\frac{5}{2}\sum_{k=1}^{N}\frac{\left( -1\right) ^{k-1}}{k^{3}\dbinom{2k}{k}}+\sum_{k=1}^{N}\frac{\left( -1\right) ^{k-1}}{2k^{3}\dbinom{N}{k}\dbinom{N+k}{k}},\tag{4} $ letting $N\rightarrow \infty $. I adapted the computation as indicated and obtained $(1)$. Since the second term on the left vanishes, as $N\rightarrow \infty $, we get the given equality in the form
$ \sum_{n=1}^{\infty }\frac{1}{n^{2}}=2\sum_{n=1}^{\infty }\frac{ (-1)^{n-1}}{n^{2}}=3\sum_{n=1}^{\infty }\frac{1}{n^{2}\dbinom{2n}{n}}.\tag{5} $
$^{1}$One of the intermediate sums can be written as $ \sum_{k=1}^{n-1}(-1)^{k}n\left( \varepsilon _{n,k}-\varepsilon _{n-1,k}\right) =\frac{1}{n^{2}}-\frac{2(-1)^{n-1}}{n^{2}\dbinom{2n}{n}},\tag{6} $
where $ \varepsilon _{n,k}=\frac{1}{2}\frac{\left( k!\right) ^{2}(n-k)!}{k^{3}(n+k)!} =\frac{1}{2k^{3}\dbinom{n+k}{k}\dbinom{n}{k}}.\tag{7} $
Instead of dividing $(6)$ by $n$ as a step to obtain $(4)$, if we multiply by $(-1)^{n-1}$ we get $ \sum_{k=1}^{n-1}(-1)^{k+n-1}n\left( \varepsilon _{n,k}-\varepsilon _{n-1,k}\right) =\frac{(-1)^{n-1}}{n^{2}}-\frac{2}{n^{2}\dbinom{2n}{n}}.\tag{8} $ After further manipulations I got $(1)$.
Note: the LHS of $(1)$ is the diagonal sequence $c_{N,N}^{\prime }$ of the double sequence $c_{n,k}^{\prime }$ defined by the formula $5^{\prime }$ in section 6 of the mentioned article.