Let $\phi : \mathbb{R}^2\rightarrow\mathbb{R}^2$ be an isometry. Suppose $\phi$ is not surjective, that is there exists some $v \in \mathbb{R}^2$ whose fiber $\phi^{-1}(v)$ is empty. Then by the pigeonhole principle there exist $u, u' \in \mathbb{R}^2$ where $u\neq u'$ which map to the same element $\phi(u)$. But then $\phi$ is not an isometry since $d(u,u') > d(\phi(u),\phi(u))=0$.
My issue is with using pigeonhole principle for uncountable sets, which feels flawed to me.
Indeed, your problem is using the pigeonhole principle for infinite sets (not even uncountable). To wit, consider the map $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = 2x$. Since $f^{-1}(1)$ is empty, by your argument, $f$ must not be injective, which is clearly false.