another proof of the non-existence of a continuous function that switches the rational and the irrational numbers

Let $𝕁=ℝ\setminus \mathbb{Q}$ denote the irrationals.There is no continuous function $f:ℝ\to ℝ$such that $f(\mathbb{Q})\subseteq 𝕁$ and $f(𝕁)\subseteq \mathbb{Q}$.

Proof

Suppose $f$ is such a function. Since $\mathbb{Q}$ is countable, $f(\mathbb{Q})$ and $f(𝕁)$ are also countable. Therefore the image of $f$ is countable. If $f$ is not a constant function, then by the intermediate value theorem the image of $f$ contains a nonempty interval, so the image of $f$ is uncountable. We have just shown that this isn’t the case, so there must be some $c$ such that $f(x)=c$ for all $x\in ℝ$. Therefore $f(\mathbb{Q})=\{c\}\subset 𝕁$ and $f(𝕁)=\{c\}\subset \mathbb{Q}$. Obviously no number is both rational and irrational, so no such $f$ exists.