Theorem. Let $f:X\to Y$ be a continuous open map with discrete fibers, $X$ second-countable Hausdorff space, and $Y$ a Baire space and $G_\delta$ space, then $$O_f := \{x\in X : f\text{ is a local homeomorphism at }x\}$$ is a dense open subset of $X$, and so there is a dense open subset $O_f\subseteq X$ such that $f$ restricted to $O_f$ is a local homeomorphism.
This theorem implies yours since 1) continuous open image of a Baire space is a Baire space, 2) if $Y$ is a second-countable Hausdorff normal space, then by Urysohn's metrization theorem, $Y$ is metrizable, and so a $G_\delta$ space.
The proof will follow the one of theorem 2.1 in Local topological properties of countable mappings by Väisälä, the only detail needing changing is at the end.
Proof: Let $(U_n)$ be a countable basis for open non-empty $U\subseteq X$, and define $E_n = \{y\in Y : |U_n\cap f^{-1}(y)| = 1\}$ so that $E_n\subseteq f(U_n)$. We'll show that $E_n$ is closed in $f(U_n)$. Let $y\in f(U_n)\setminus E_n$, then there are distinct $x_1, x_2\in U_n\cap f^{-1}(y)$, and we can take disjoint open $V_1, V_2$ with $x_k\in V_k\subseteq U_n$. Then $y\in f(V_1)\cap f(V_2)\subseteq f(U_n)\setminus E_n$ and so $f(U_n)\setminus E_n$ is open. Since $f$ has discrete fibers, $f(U) = \bigcup_n E_n$. Since $Y$ is a $G_\delta$ space, $f(U_n)$ is an $F_\sigma$-set for every $n$, and so every $E_n$ is an $F_\sigma$-set. Since open subspace of a Baire space is a Baire space, $f(U)$ is a Baire space, and so there is $n$ such that $\text{int}(E_n)\neq \emptyset$. Then $f^{-1}(\text{int}(E_n))\cap U_n$ is a non-empty subset of $O_f\cap U$. It follows that $\overline{O_f} = X$. $\square$
