I tried to solve the following question from Munkres (for reference: chapter 81, exercise 3 a, page 493)
Let $p:X \to B$ be a covering map and $G$ its group of covering transformations. Then $G$ acts properly discontinuous on $X$.
So the action is properly discontinuous if for every $x \in X$ there is an open neighborhood such that if $g(U) \cap U \neq \emptyset$ then $g$ is the identity homeomorphism. Considering $x \in X$, there is an evenly covered neighborhood of $p(x)$, say $V$. Hence we can write $p^{-1}(V)$ as disjoint union of open subsets $U_\beta$ in $X$ which are homeomorphic under $p$ to $V$. Consider $U$ to be the open neighborhood containing $x$. I tried to prove that for $g$ not the identity homeomorphism, we must have that $g(U) \cap U$ is empty, but I could not do this. Any hints would be appreciated.
EDIT I think I could answer my question using Bass' answer. Any remarks/hints on how to improve my answer would be appreciated (I was not sure if I had to post my answer as an actual answer, or if I should have edited it in this question. I chose the former, see my answer below)