Let $f: U\rightarrow \mathbb{C}$ is analytic, and $f'(z)\neq 0$ for all $z \in U$ , then $f$ is open mapp in U. (Send open sets to open sets).
I'm not quite sure how to write the arguments. But, using the inverse function theorem, then there exists an open neighborhood $U'$ for $z_0\in U$ and $V'$ for $f(z_0)$ where $g:V' \rightarrow U'$ is the inverse of $f$ and $g$ is analytic.
For $A\subset U'$, an open set, since $g$ is analytic then g is continuous, then the inverse image $g^{-1}(A)$ is an open set. But, there exists a set $B\subset V'$ such that $B=g^{-1}(A)$.