Proof that sequence converges to staircase function

In the following I assume you are interested in point-wise convergence.

Notice that points of the form $k\pi$ are fixed points of $f$, and so are fixed by every function in your sequence $\{ f_n \}_n$. Moreover, notice that $f(x)=-f(-x)$, and so this also holds for every function in the sequence. As a consequence, it is enough to focus on points $x\geq 0$, and the rest follows by symmetry. Moreover, by an argument similar to the one above, it can easily be seen that it is sufficient to study the sequence in the interval $[0,\pi]$.

It is enough to prove that, if $x=(2k+1)\pi+y$, with $-\pi<y<\pi$, then $f^n(x)\to (2k+1)\pi$, which can be easily shown by working in $[0,\pi]$.

Note that for $k \in \mathbb{Z}$;

$\lim_{n \rightarrow \infty}f_n(k\pi) = k\pi$

Now there are two remaining cases where the domain can belong

Case 1) $x \in ((2k-1)\pi,2k\pi)$

Then $(2k-1)\pi < f(x) < x$; Hence $f_n(x)$ is a decreasing sequence bounded below by $(2k-1)\pi$; Thus $f_n(x)$ must converge to some $p_x$ in $((2k-1)\pi,2k\pi)$ that must be a fixed point; thus we must have $p_x = (2k-1)\pi$

Case 2) $y \in (2k\pi,(2k+1)\pi)$

Note that in this case $ y < f(y) < (2k+1)\pi$; Hence $f_n(y)$ is a increasing sequence bounded above by $(2k+1)\pi$; by similar reasoning to Case 1) we have that $f_n(y)$ converges to $(2k+1)\pi$