How to prove the following problem:
It is given Hausdorff space $X$ and finite group $G$ (with neutral $e$) that is acting freely on $X$. For $g\in G$, $\overline{g}:X\rightarrow X$ is homeomorphism.
a) Prove that for every $x\in X$ exists an open neighborhood $U$ such that for every $g\in G$, $g\neq e,$ the property $U\cap \overline{g}(U)=\emptyset$ holds.
b) Prove that for each neighbourhood $U$ from part a) we have $\overline{g_1}(U)\cap\overline{g_2}(U)=\emptyset$ for every two distinct elements $g_1, g_2\in G$.
c) Prove that $\pi : X\to X/G$ is covering map.
Any kind of help is welcome.
Let $x \in X$, $g\in G - \{e\}$. As $\bar gx \ne x$, by Hausdorffness there are open neighbourhoods $U_g$ of $x$, $V_g$ of $\bar gx$ such that $U_g \cap V_g \ne \emptyset$. Let $U = \bigcap_{h \in G-\{e\}} (U_h \cap \bar h^{-1}V_h)$. Then $U$ is an open neighbourhood of $x$ and for $g \in G$ we have $$ \bar g U \cap U \subseteq \bar g\bar g^{-1}V_g \cap U_g = \emptyset. $$ For b) note that $$ \bar g_1 U \cap \bar g_2 U = \bar g_1 (U \cap \overline{g_1^{-1}g_2}U) = \bar g_1\emptyset = \emptyset. $$ For c), let $\pi(x) \in X/G$, let $U$ as in b). Then we have that $\pi(U)$ is open in $U$ as $\pi^{-1}\pi U = \bigcup_{g\in G} \bar g U$ is open, hence an open neighbourhood of $\pi(x)$. Moreover, restricted to each $\bar g U$, $\pi$ is a homeomorphism onto $[U]$, therefore, as the $\bar g U$ are disjoint, $\pi$ is a covering map.