I am working on proving that if $q:E\rightarrow X$ is a covering map, and $X$ is Hausdorff, then so is $E$.
The answer to this question: Domain is Hausdorff if image of covering map is Hausdorff suggested choosing any points $e_1,e_2\in E$ and considering the cases $q(e_1)=q(e_2)$ and $q(e_1)\neq q(e_2)$.
In the case $q(e_1)\neq q(e_2)$, I argue as follows. Since $q(e_1),q(e_2)\in X$, and $X$ is Hausdorff, we can find disjoint open $U_1$ and $U_2$ with $q(e_1)\in U_1$ and $q(e_2)\in U_2$. Then, $q^{-1}(U_1)$ and $q^{-1}(U_2)$ are open and they must be disjoint, because if $x\in q^{-1}(U_1)\cap q^{-1}(U_2)$ then $q(x)\in U_1$ and $q(x)\in U_2$, which cannot be the case, these sets should be disjoint.
This seems like it works to me, but it doesn't use the properties of covering maps, in fact this would work if $q$ is any quotient map, I think? What am I missing?
As iwriteonbananas said, you'll use the covering map property in the other case $q(e_1)=q(e_2)$.
Your reasoning for the case $q(e_1)\ne q(e_2)$ is correct. If both points have distinct images, then disjoint neighborhoods of these images pull back to disjoint neighborhoods around the points $e_1$ and $e_2$. This works for any continuous map.
In the case $q(e_1)=q(e_2)=x$, we can find disjoint neighborhoods of $e_1$ and $e_2$ by looking at the preimage of an evenly covered neighborhood of $x$. Being a local homeomorphism is not enough as the map $(I\times\{0,1\})\,/\,((s,0)\sim (s,1)\forall s>0)\:\to\:I, \;(s,t)\mapsto s $ demonstrates.