Supremum over real interval equals supremum over rational 'interval' for continuos functions?

Question

I've recently read the following reasoning in a paper:

the mapping $x\mapsto \sup_{s\in[a,b]} f(sx)$, $\mathbb{R} \to \mathbb{R}$, is Borel-measurable, since the supremum over $s\in[a,b]$ equals the supremum over $s\in[a,b]\cap\mathbb{Q}$ due to the continuity of $f$ and the fact that the supremum of a countable collection of measurable functions is itself measurable.

While I understand the last part concerning the measurability of a countable supremum, I was not able to find resources that prove that $$ \sup_{s\in[a,b]} f(s) = \sup_{s\in[a,b]\cap\mathbb{Q}} f(s) $$ for a continuous function $f$, which was probably used in the first part.
Can anyone help me out with that?

Answer 1

Clearly $$\sup_{s\in[a,b]} f(s) \geq \sup_{s\in[a,b]\cap\mathbb{Q}} f(s)$$ Suppose by way of contradiction that $$M=\sup_{s\in[a,b]} f(s) > \sup_{s\in[a,b]\cap\mathbb{Q}} f(s)=m$$ and let $\varepsilon = \frac{M-m}2$. The exists an $x\in [a,b]$ such that $|M-f(x)|<\varepsilon.$ There exists a sequence of of rational numbers $a\leq x_n\leq b$ such that $x_n\to x$. Since $f$ is continuous, $f(x_n)\to f(x)>m+\frac{\varepsilon}2$ as $n\to\infty$.

But this is absurd, since $f(x_n)\leq m$ for every $n$.