Question:
Does a Lipschitz continuous surjection $f:\mathbb R \setminus \mathbb Q \to \mathbb Q$ exist?
Without the restriction on $f$ being Lipschitz continuous the answer is yes: Give an example of a function $h:\mathbb{R}\backslash\mathbb{Q}\rightarrow\ \mathbb{Q}$.
The intuitive reason why I restrict $f$ to be Lipschitz continuous and not only continuous is that I expect a Lipschitz continuous function on a dense subset of an interval $I$ to have similar properties as if it was continuous on $I$.
The answer is no.
Proof: Suppose to the contrary that there was such a Lipschitz continuous surjection $f:\mathbb R \setminus \mathbb Q \to \mathbb Q$. Then by Kirszbraun theorem it could be extended into a function $g:\mathbb R \to \mathbb R$ the is Lipschitz continuous as well. Consider any $a,b\in \mathbb R \setminus \mathbb Q$ such that $f(a)\neq f(b)$. Then since $g$ is continuous (on $\mathbb R$), $$ [g(a),g(b)] \subset g[a,b] = f([a,b]\setminus \mathbb Q) \,\cup\, g([a,b]\cap \mathbb Q), $$ where $[g(a),g(b)]$ is an uncountable set, whilst both $f([a,b]\setminus \mathbb Q)\subset \mathbb Q$ and $g([a,b]\cap \mathbb Q)$ are both countable. $\tag*{$\Box$}$