Romanian Mathematical Olympiad

Since $f$ is monotonic, the lateral limits at each point exist and are finite. Denote $f(x_0^{-})$, respectively $f(x_0^{+})$, the limit from the left, respectively from the right at $x_0$. Then, $f$ being increasing, $f(x_0^{-})\le f(x_0)\le f(x_0^{+})$ for every $x_0\in \mathbb{R}$.

Suppose now that $f$ is discontinuous at some point $a$. Then there exists $A,B$ such that $f(a^{-})<A<B<f(a^{+})$. It follows, since $f$ is increasing, that $f(x)\le A$ for all $x<a$ and $f(x)\ge B$ for all $x>a$. Using again the monotony, $f(f(x))\le f(A)$ for all $x<a$ and $f(f(x))\ge f(B)$ for all $x>a$. But, this implies $f(f(a^{-}))\le f(A)<f(B)\le f(f(a^{+}))$, which would mean that $f\circ f$ is discontinuous at $a$, a contradiction.