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

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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 $y\in U^c$?

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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 $y$.