A continuous function with rational values is constant

Idea. A continuous function cannot skip any value between two of its values. But the rational numbers have gaps: between two rationals there are irrational numbers.

Proof. Suppose $f(a)\neq f(b)$ for some $a,b\in[0,1]$, say $f(a)<f(b)$. The interval $\bigl(f(a),f(b)\bigr)$ contains an irrational number $y$ (for example $f(a)+\frac{\sqrt2}{2}\,\bigl(f(b)-f(a)\bigr)$, which lies strictly between since $0<\frac{\sqrt2}2<1$, and is irrational because $f(b)-f(a)$ is a non-zero rational).

By the intermediate value theorem on the interval between $a$ and $b$, there is $c$ with $f(c)=y$. But $f(c)$ is rational and $y$ is not. Contradiction.

Conclusion. $f(a)=f(b)$ for all $a,b$, so $f$ is constant.

Remark. The same argument shows that a continuous function with values in $\mathbb{Z}$ is constant, and more generally that a continuous function on an interval cannot take only values in a countable set unless it is constant.