I am trying to prove the following conjecture:
Let $\pi : E\to B$ be a covering map. If $E$ is locally compact Hausdorff, then so is $B$.
(The converse is known to be true, cf. Exercise 6 in Section 53 of Munkres' Topology (2e)). It is not very hard to prove that $B$ is at least locally compact in the weak sense (every point $b\in B$ has a compact neighborhood) since $\pi$ is a continuous open map. So the issue I'm having lies with Hausdorffness of $B$.
Here is my attempt so far: let $b, b'\in B$ be distinct points. Choose an open set $V\subseteq B$ containing $b$ that is evenly covered by $\pi$, say $\pi^{-1}(V) = \bigcup_{\lambda\in\Lambda} U_\lambda$. If $b' \in V$, then since $E$ is Hausdorff and each map $\pi|_{U_\lambda} : U_\lambda \to V$ is a homeomorphism, it follows that $V$ is Hausdorff, and hence $b$ and $b'$ can be separated by disjoint $V$-open sets. If instead $b' \notin \overline{V}$, then $b$ and $b'$ can be separated by the open sets $V$ and $B\setminus\overline{V}$.
There's one remaining case: suppose instead that $b'$ lies on the boundary $\partial V$. Pick any point $e\in \pi^{-1}(\{b\})$ and corresponding leaf $e\in U_{\lambda_0}$ from the even cover of $V$. Since $E$ is locally compact Hausdorff, there exists a precompact open set $W \subseteq E$ satisfying $e\in W\subseteq \overline{W} \subseteq U_{\lambda_0}$. I would like to show that $\pi(\overline{W})$ is closed in $B$. If this is the case, then $b$ and $b'$ can be separated by the disjoint open sets $\pi(W)$ and $B \setminus \overline{\pi(W)}$.
We know at least that $\pi^{-1}(\pi(\overline{W})) = \bigcup_{\lambda\in\Lambda} (\pi|_{U_\lambda})^{-1}(\pi(\overline{W})) = \bigcup_{\lambda\in\Lambda} \overline{(\pi|_{U_\lambda})^{-1}(\pi(W))} \subseteq \overline{\pi^{-1}(\pi(W))}$ since each slice $(\pi|_{U_\lambda})^{-1}(\pi(\overline{W}))$ is compact and hence closed in $E$. My intuition tells me that the reverse inclusion is true too, but this is where I'm stuck. Is there a way to prove the reverse inclusion here? Or is it possible that $B$ is not Hausdorff after all?
