The problem reminds me of Euler's (very famous) heuristic proof of the Basel problem, as described here. Inspired by that, we can concoct a similar heuristic explanation for this problem. In fact, our heuristic will be detailed enough that it actually yields a rigorous proof using the Weierstrass factorization theorem.
First, rewrite the fixed point equation $x=\tan x$ as $$\frac{\sin x}{x}=\cos x,$$
which (a la Euler) can be written as a product:
$$
\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(n-\tfrac12)^2\pi^2}\right).
$$
Now (just as if these were polynomials) we want to move everything over to the left side and factor. When we consider
$$
\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)-\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(n-\tfrac12)^2\pi^2}\right),
$$
we see that the constant term $1$ cancels out, and there is no coefficient of $x$ (since all powers of $x$ appearing are even) so that leaves the $x^2$ coefficient:
$$
\sum_{n=1}^{\infty}\frac{1}{(n-\tfrac12)^2\pi^2}-\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3},
$$
where we calculated sum involving $n-\tfrac12$ from the sum involving $n$ by thinking of it as $4$ times the odd terms in the $n$ sum, and calculated the odd sum as the whole sum minus the even terms (which contribute $\tfrac14$ times the whole sum). (We will use this trick once more at the very end...)
Therefore we have the factorization
$$\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)-\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(n-\tfrac12)^2\pi^2}\right)=\frac{x^2}{3}\prod_{n=1}^{\infty}\left(1-\frac{x^2}{u_n^2}\right).$$
Thus we will be able to compute $\sum_{n=1}^{\infty}\frac{1}{u_n^2}$ by computing the $x^{4}$ coefficient of the left side.
Therefore we find that
$$
\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{u_n^2}=\sum_{1\leq n<m}\left[\frac{1}{n^2m^2\pi^4}-\frac{1}{(n-\tfrac12)^2(m-\tfrac12)^2\pi^4}\right].
$$
Since in general
$$
\sum_{n<m}a_na_m=\frac{\Bigl(\sum_{n}a_n\Bigr)^2-\sum_n a_n^2}{2},
$$
we can apply this trick separately to both terms to write them in terms of $$\sum_{n=1}^{\infty}\frac{1}{n^4\pi^4}=\frac{1}{90}\qquad\text{and}\qquad \sum_{n=1}^{\infty}\frac{1}{(n-\tfrac12)^4\pi^4}=\frac{1}{6},$$
(where again we get the $n-\tfrac12$ sum from the $n$ sum using the $\textrm{odd}=\textrm{all}-\textrm{even}$ trick) and after some straightforward manipulations we obtain that
$$
\sum_{n=1}^{\infty}\frac{1}{u_n^2}=\frac{1}{10}.
$$
