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, $f(x_0^-) \leq f(x_0) \leq f(x_0^+)$ 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 $f(x) \leq A$ for all $x < a$ and $f(x) \geq B$ for all $x > a$. Using again the monotony, $f(f(x)) \leq f(A)$ for all $x < a$ and $f(f(x)) \geq f(B)$ for all $x > a$. But, this implies $f(f(a^-)) \leq f(A) < f(B) \leq f(f(a^+))$, which would mean that $f \circ f$ is discontinuous at $a$, a contradiction.