Proof of the quotient map $\pi : X\to X/G$ is a covering map only if the action of $G$ is properly discontinuous. The following is a theorem from Munkres' Topology and there's a part in the proof of the theorem that I don't understand. I've written the part in bold. $X/G$ is the orbit space obtained from $X$ by means of the equivalence relation $x \sim g(x)$ for all $x\in X$ and all $g \in G$, and $e$ is the identity element of $G$.
Theorem 81.5. Let $X$ be path connected and locally path connected; let $G$ be a subgroup of the group of homeomorphisms of $X$. The quotient map $\pi : X \to X/G$ is a covering map if and only if the action of $G$ is properly discontinuous. In this case, the covering map $\pi$ is regular and $G$ is its group of covering transformations.
Proof: We suppose now that $\pi$ is a covering map and show that the action of $G$ is properly discontinuous. Given $x\in X$, let $V$ be a neighborhood of $\pi(x)$ that is evenly covered by $\pi$. Partition $\pi^{-1}(V)$ into slices; let $U_\alpha$ be the slice containing $x$. Given $g\in G$, with $g\neq e$, the set $g(U_\alpha)$ must be disjoint from $U_\alpha$, for otherwise, two points of $U_\alpha$ would belong to the same orbit and the restriction of $\pi$ to $U_\alpha$ would not be injective. It follows that the action of $G$ is properly discontinuous.
I have trouble figuring out the bolded part. Letting $y\in U_\alpha \cap g(U_\alpha)$, I don't see how this leads to a problem in the case $g(y)=y$, since then we don't get two different points belonging to the same orbit. In the most extreme case, what problem do we have if $g$ is identity restricted to $U_\alpha$? I would greatly appreciate it if anyone could explain this line to me.
 A: I actually just figured out the answer to this question. I post this as an answer for anyone curious to see. The key was in the next step of the proof, which is why I wasn't able to see it, I think Munkres just assumed it was quite obvious. 
So clearly the only problem arises when there is a $g\in G$ that is not the identity, such that $g$ sends points in $U_\alpha$ identically. But this cannot be the case, because $G$ here becomes the group of covering transformations. Certainly any $g\in G$ is a covering transformation, for $\pi \circ g=\pi$ because the orbit of $g(x)$ equals the orbit of $x$. On the other hand, let $h$ be a covering transformation with $h(x_1)=x_2$, say. Because $\pi \circ h=\pi$, the points $x_1$ and $x_2$ map to the same point under $\pi$; therefore there is an element $g\in G$ such that $g(x_1)=x_2$. Now by the uniqueness of covering equivalences, $g=h$.
And by the same reasoning, if $g\in G$ sends a point of $U_\alpha$ to the same point, then it must be $e$. So we can ignore this case.
