# Subgroups of covering space actions

Suppose that $G$ is a covering space action such that for each $x \in X$ there is a neighbourhood $U_x$ such that all the images of $g(U_x)$ for varying $g \in G$ are disjoint. In other words $g_1(U_x) \cap g_2(U_x) \neq \emptyset \implies g_1 = g_2$.

Suppose that $H \subset G$, it is true that $H$ is also a covering space action that satisfies the above condition?

See Hatcher page 72.

Hatcher states (without proof) in Exercise 24 of this section that each subgroup $H \subset G$ of a covering space action of a group $G$ on a space $X$ induces a composition of covering spaces $X \to X/H \to X/G$.
Because $X \to X/G$ is normal (see Proposition 1.40a), one can prove that $X \to X/H$ is normal. Indeed, the points of $X/G$ are the orbits $Gx$ of points $x \in X$, so for a point $Hx \in X/H$, there is a natural point $Gx \in X/G$ from the natural map $X/H \to X/G$. Two lifts $x$ and $x'$ of $Hx \in X/H$ are also lifts of $Gx \in X/G$, and so there is a deck transformation taking $x \mapsto x'$ by normality of $X \to X/G$, so $X \to X/H$ is normal.
So, $X \to X/H$ is a normal covering space given by the covering space action of $H \subset G$. I don't know if $H$ necessarily satisfies the condition you've cited in the beginning of your question, but the important consequence of this condition -- that $X \to X/H$ is normal -- is satisfied. That exercise implies that the other parts of Proposition 1.40 are also satisfied for a subgroup $H \subset G$.