There is an exercise problem in Hatcher's Algebraic Topology book asking to show that if $p:\tilde{X}\rightarrow X$ is a covering space with $p^{-1}(x)$ finite and nonempty for all $x\in X$, then $\tilde{X}$ is compact Hausdorff iff $X$ is compact Hausdorff.
I've managed to show each part of the statement except for $\tilde{X}$ Hausdorff $\Rightarrow X$ Hausdorff. The problem I encountered was that taking $x\neq y$ in $X$ and $\tilde{U},\tilde{V}$ disjoint open sets in $\tilde{X}$ with $p^{-1}(x)\subset\tilde{U}$ and $p^{-1}(y)\subset \tilde{V}$ (I'm using the assumption that $\tilde{X}$ is compact Hausdorff, thus normal), I could not use these to get disjoint open sets in $X$ separating $x,y$.
Best Answer
Take distinct $y_1 \neq y_2$ in $X$. Then $F_1 = f^{-1}[\{y_1\}]$ and $F_2 = f^{-1}[\{y_2\}]$ are disjoint, finite and non-empty subsets of $\tilde{X}$. By Hausdorffness we can find disjoint open sets $U_1$ and $U_2$ around $F_1$ and $F_2$. Let $O_1$ be an evenly covered neighbourhood of $y_1$, and $O_2$ one of $y_2$.
Now, $f^{-1}[O_1] = \cup_{i=1}^k V_i$, where the $V_i$ are disjoint and open and $f|V_i$ is a homeomorphism between $V_i$ and $O_1$ for all $i$; define $V'_i = V_i \cap U_1$, and define $O'_1 = \cap_{i=1}^k f[V'_i]$, which is open, as a finite intersection of open sets.
Similarly we define $W'_i$ from the inverse image of $O_2$ and $U_2$, and the intersection of their images is called $O'_2$.
These $O'_1$ and $O'_2$ are the required disjoint open neighbourhoods as can be easily checked.