Let $G$ be a finite group that acts on a topological space $X$ (path connected) freely. Then prove that $q:X \to X/G$ is a covering map.
I am unable to think about this. Please provide me with some hints.
My Trying:
Let $[x] \in X/G$. Choose $x \in X$ along with an open set $\tilde{U}$ in $X$. Since here the quotient map $q:X \to X/G$ is an open map we have $q(\tilde{U})$ as an open set in $X/G$ containing $[x]$. Now it remains to show that $q(\tilde{U})$ is an evenly covered open nbd. of $[x]$.
Now we have, $q^{-1}(q(\tilde{U})) = \bigcup_{g \in G} g \tilde{U}$, i.e., union of the left translates of $\tilde{U}$ in $X$ by $G$. Since left translation gives a homeomorphism from $X$ to $X$, it induces a homeomorphism between $U$ and $g\tilde{U}$ for each $g \in G$.
It remains to show that $g_1 \tilde{U} \cap g_2 \tilde{U} = \phi$ for any two distinct $g_1$ and $g_2$ in $G$. For this we use the fact that the action is free. Let $\alpha \in g_1 \tilde{U} \cap g_2 \tilde{U}$ then $\alpha = g_1u = g_2 u'$ for some $u,u' \in \tilde{U}$.
But from here I am unable to arrive at a contradiction.
Are the above arguments correct? Please correct me if I am wrong and help me on the last part of the proof.