Here's a very detailed proof.
Let's say we have a continuous map $f:X \to Y$ of topological spaces of which we know:
- $f$ is a local homeomorphism, that is for every $p \in X$ exist the open subsets $U \subseteq X$, $V \subseteq Y$ with $p \in U$ and such that $f_{|U}:U \to V$ is a homeomorphism
- $f$ is bijective, that is there is an inverse map $f^{-1}:Y \to X$
In order to prove that $f$ is a homeomorphism we need to prove that $f^{-1}$ is continuous.
So, let $U' \subseteq X$ an open set and $V' = (f^{-1})^{-1}(U') = f(U')$. For each $p \in V'$ let $U_p$, $V_p$ as above (i.e. $f_{|U_p}: U_p \to V_p$ is homeomorphism), then $ V' \cap V_p = f_{|U_p}(U' \cap U_p) $ is open because $f_{|U_p}$ is an homeomorphism (and therefore an open map). Furthermore $V'= \cup_{p \in V'} V' \cap V_p$ is open, as union of open sets.