Let $f:\Omega \to \mathbb{C}$ be 1-1 and analytic on open set $\Omega \subset \mathbb{C}$. If $z_n \in \Omega \to \partial \Omega$ then prove $f(z_n) \to \partial f(\Omega)$ in the sense that $f(z_n)$ eventually lies outside each compact subset of $f(\Omega)$. This property means $f$ is proper.
I thought of proof by contradiction, if $\{f(z_n)\}$ is inside a compact set then there is a limit point inside $f(\Omega)$, does this show any contradiction?
Recall the open mapping theorem, which tells you, that non-constant analytic maps on connected domains are open maps. Using this and the injectivity of your map, you can prove that $f$ is an open map. Thus, the inverse map is continuous and hence, your map is proper (by the continuity of the inverse map, the preimage of compact sets are compact).
Added: Write $\Omega= \bigcup_{n\in I} U_n$ with $U_n$ being the connected components of $\Omega$. Note that $f$ restricted to any $U_n$ is an open map as the $U_n$ are open sets. If $V\subseteq \Omega$ is an open set, then we get by injectivity
$$ f(V) = \bigcup_{n\in I} f(U_n \cap V), $$
which is open.
Alternative proof: Injective analytic maps have nonzero derivative (Why do injective holomorphic functions have nonzero derivative) and thus by the inverse function theorem, $f$ is a local (analytic) diffeomorphism on its image. By injectivity $f$ is a global (analytic) diffeomorphism on its image and is therefore a proper map.
