Let $(X,d)$ a complete metric space. We have a continuous function $f:X\to\,\mathbb{Q}$ , the rationals. Show that there exists an open set $G$ in $X$ such that $f$ is constant on $G$.
My proof relies on a theorem (not very well known) which has a rather difficult proof, saying that if $X=\bigcup_{n}A_{n}$ with all $A_{n}$ closed, then there is some $A_{n}$ containing an open neighborhood $S(x,\epsilon)$.
Let $\left\{q_{1},q_{2},.... \right\}$ be the sequence of the rational numbers.
Take $A_{n}=f^{-1}\left\{q_{n} \right\}$. Then by continuity each $A_{n}$ is closed.
And it is clear that $X=\bigcup_{n}A_{n}$. By invoking the theorem we get that there is an open neighborhood $S(x,\epsilon)$ in at least one $A_{n}$ and hence $f(x)=q_{n}$ constant on this open neighborhood.
So, the question is, can we find a direct, self-contained proof of the result, without referring to this theorem? I believe it would be something hard to do.
The link of the theorem is https://sites.math.washington.edu/~morrow/336_14/papers/alana.pdf
The preimage of a rational number $q$ is a closed subset of $X,$ and by your assumption it has empty interior, so is nowhere dense, so $X$ is a countable union of nowhere dense sets...