Q:

Prove that continuous functions mapping irrationals to rationals must be constant

Let $f\colon[0,1] \to \mathbb{R}$ be a continuous function 
such that any irrational number is mapped to a rational number. Then $f$ must be a constant.
Well, the context isn't that much, I was reading some interesting function constructions on Hardy's book. Then raised this question. 
As the solution suggest it is an elementary exercise. But definitely interesting.

A:

For convenience, let $\mathbb{I} = [0,1]\cap (\mathbb{R}\setminus\mathbb{Q})$, which is uncountable, and $\mathbb{J} = [0,1]\cap\mathbb{Q}$, which is countable.

$f(\mathbb{I}) \subseteq \mathbb{Q}$ is countable.
$f(\mathbb{J})$ is countable since $\mathbb{J}$ is.

So $f([0,1])=f(\mathbb{I})\cup f(\mathbb{J})$ is countable.
Now, $[0,1]$ is an interval, and the image of an interval by a continuous function is an interval. So $f([0,1])$ is a countable interval, which means...

