Somehow I seem to be missing something, since I expected to find an answer to this question somewhere but that is not so.
Suppose a topological group $G$ acts continuously on some $G$-space $X$. If the action is a covering space action (i.e. each $x\in X$ admits a neighborhood $U$, such that $gU\cap U=\emptyset$ for each $g\in G\setminus\lbrace e\rbrace$) is it necessarily true, that $G$ must be a discrete group?
Thanks in advance for any insights!
First, note that for any $x \in X$, one may find a neighborhood $U_x$ containing $x$, which implies $gU_x \cap U_x = \varnothing$. Hence $gx \neq x$, so $G$ acts freely.
Since $\pi \colon X \to X / G$ is a covering map, each fiber $\pi^{-1}([x]) = Gx$ is a discrete subset of $X$. Let $\iota_x \colon G \to G \times X$ be the inclusion $g \mapsto (g,x)$. Since the action $\rho \colon G \times X \to X$ is continuous, the map $\rho_x = \rho \circ \iota_x \colon G \to Gx$, which is given by $\rho_x(g) = gx$, is also continuous.
Since $G$ acts freely, $\rho_x$ is injective, and thus a continuous bijection. The inverse is automatically continuous, as a map from a discrete space. Hence $G \cong Gx$, which implies $G$ is also discrete.