Covering space Hausdorff implies base space Hausdorff 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 $x,y$.
 A: Take $x \neq y$ in $X$ and choose $x' \in \tilde X$ mapping to $x$.  Choose a neighborhood $U'$ of $x'$ and neighborhoods $V_{y'}$ for each $y' \in p^{-1}(y)$ such that none of these neighborhoods intersect.  Let $U$ be the image of $U'$ in $X$.  Define $V$ by pushing the $V_{y'}$ to $X$ first and then intersecting.
Show that $U$ and $V$ are neighborhoods of $x$ and $y$ that don't intersect.
Edit:  Oops, forgot to say you'll want to choose the $V_{y'}$ so that no two intersect.  I've edited my answer above to reflect this.  You can do this by proving the following little lemma (if you don't already know it) and applying it to $\{x'\} \cup p^{-1}(x)$.
Lemma: Let $Y$ be Hausdorff and $\{v_1, \ldots, v_n\}$ a set of finitely many distinct points in $Y$.  Then there exist neighborhoods $V_i$ of $v_i$ such that no two of the $V_i$ intersect.
A: A space $X$ is Hausdorff if and only if the diagonal is closed in $X \times X$. Now suppose $\tilde X$ is Hausdorff, so the diagonal $\Delta_{\tilde X}$ is closed in $\tilde X \times \tilde X$. Remark that $p \times p : \tilde X \times \tilde X \to X \times X$ is again a finite covering map, hence it is a closed map; and the surjectivity of $p$ implies that $(p\times p) (\Delta_{\tilde X}) = \Delta_{X}$, so $\Delta_X$ is closed in $X \times X$ and therefore $X$ is Hausdorff.
A: 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.
A: For $x\neq y$ in $X$, we have evenly covered neighbourhoods $U_{x}, U_{y} $ of $x,y$ respectively, and if these neighborhoods are disjoint, we are done. If both $x,y$ are in some same evenly covered neighborhood $U \subset X$, then also, we are done, as this neighborhood is the homeomorphic image of an open set of the covering space $X^{'}$ that is Hausdorff.
So, we assume $U_{x},U_{y}$ are intersecting, with $x$ and $y$ both not in one of them. The basic property of covering spaces then implies that $x,y$ both have an equal number, $n$ (say), of preimages. Further, every element of $U_{x} \cup U_{y}$ has the same number $n$, of preimages in $X^{'}$.
Let the preimages of $x$ in $X^{'}$ be labelled $x_{1},..,x_{n}$, and those of $y$ be labelled $y_{1},..y_{n}$. With the procedure of the previous answer, you can choose neighborhoods in $X'$ around each of these $2n$ points so that each of these neighborhoods are disjoint from another, and all of them get mapped homeomorphically to $U_{x} \cup U_{y}$ by the covering map. 
Choose any $(n+1)$ neighborhoods of these $2n$ neighborhoods. Then at least one of them has to contain a preimage of $x$, and at least one of them has to contain a preimage of $y$. Of the (n+1) neighbourhoods chosen, collect all of them containing a preimage of $x$, push to $X$, and then intersect. Call this set $V_{x}$. Choose the remaining, which are neighbourhoods of preimages of $y$,  push to $X$, and intersect. Call this $V_{y}$.
If we have  $a\in V_{x}\cap V_{y} \subset U_{x}\cup U_{y}$ , then $a$ has to have $(n+1)$ preimages in $X'$ which is a contradiction. 
