Q:

If $X$ is Hausdorff, then so is $E$

Let $q:E \to X$ be a covering map.  If $X$ is Hausdorff, then so is $E$.
OK, suppose $X$ is Hausdorff and let $x,y \in E$ with $x\neq y$. Let $V$ denote the evenly covered neighbourhood for $q(x)$, and let $U$ be the sheet of the covering over $V$ which contains $x$. Then $q$ restricts to a homeomorphism on $U$. Then since $X$ is Hausdorff, so is $V$ and therefore $U$. If $y \in U$, then we're done.  If $y \in U^c$.....
I'm stuck here.  How do we find separating neighbourhoods of $x,y$ when $y\in U^c$? 

A:

If $q(x) = q(y)$ and $x \neq y$, then $x$ and $y$ must be in separate sheets since they are both in the fiber over $q(x)= q(y)$, and you are done.   If they were in the same sheet, then you would have two elements of the same sheet mapping to the same element, contradicting injectiveness of $q$ when restricted to each sheet. 
Otherwise $q(x) \neq q(y)$ and since $X$ is Hausdorff, there exist disjoint open sets $V_1$ and $V_2$ containing $q(x)$ and $q(y)$. Now try to use the fact that $q$ is a covering map to get your neighborhoods of $x$ and $y$.

