If $f \circ f$ continuous prove $f$ continuous 
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ strictly increasing so that
  $f \circ f$ is continuous. Prove $f$ is continuous.

I can prove this using sequences, but it's quite tedious.
My question is: Does anyone know how to prove this using some kind of more general result (a theorem or something)?

A strictly increasing function $f:\Bbb R\to\Bbb R$ is continuous if and only if it has no jump discontinuity, which is the case if and only if $f[\Bbb R]$ is order-convex. Thus, if $f$ is not continuous, there are $a,b,c,d\in\Bbb R$ such that $a\le b<c\le d$, $(c,d)\cap f[\Bbb R]=\varnothing$, and $a,d\in f[\Bbb R]$. Let $u,v\in\Bbb R$ be such that $f(u)=a$ and $f(v)=d$. Then 
$$(f\circ f)\big[[u,v]\big]=f\big[[a,d]\big]$$
is not convex. Since $f\circ f$ is strictly increasing, $(f\circ f)[\Bbb R]$ is not convex, and hence $f\circ f$ is not continuous.
