I'm in trouble with this (maybe simple) exercise:
Let $X$ be an Hausdorff topological space and let $G$ a finite group of automorphism of $X$. Suppose that $G$ acts on $X$ without fixed points (i.e. $\forall \,\psi \in G \setminus \{id\}$ then $\psi(x) \neq x \quad \forall \, x \in X$). Then $p \colon X \to X/G$ together with $X$ is a cover space of $X/G$ where $p$ is the canonical projection.
My first idea is to prove that $G$ is properly discontinuous. Indeed, $G$ is a finite group, so we suppose $G= \{g_0,\dots,g_n\}$. Let $x_0 \in X$ and now we consider its orbit under the action of $G$, say $O_{x_0} =\{x_0,x_1, \dots, x_n\}$; $X$ is Hausdorff then there exist $V_i$ disjoint neighborhoods of $x_i$ respectively ($V_i \cap V_j= \emptyset \, \, \forall i \neq j$). Now, for example we take $g_i$ such that $g_i(x_0)=x_i$; we have to show that exits a neighborhood of $x_0$, say $U_i$, such that $g_i(U_i) \subset V_i.$ I think $U_i = V_0 \cap g_{i}^{-1}(V_i)$ could work, but I don't know exactly why (any ideas?). Then we call $U= \bigcap_{i=0}^{n} U_i$. Thus, theoretically, $U \subset V_0$ and $g_i(U) \subset V_i$ and so $U \cap g_i(U) = \emptyset \, \, \forall \,i$.
Now, if I knew $X$ is connected and locally arcwise-connected I would conclude the proof. So, how to prove the statement without knowing $X$ is connected and locally arcwise-connected? I think $p$ is an open map, is this information useful?
So the setup is as you propose:
$x_0 \in X$, $G = \{g_0, \ldots,g_n\}$ with $g_0 = \mathrm{id}_X$, $x_i = g_i(x_0)$ for $i \in \{0,\ldots,n\}$.
If you find a $U_0$ such that $x_0 \in U_0$ and such that $\{g_i[U_0]: i =0,\ldots n\}$ are pairwise disjoint then $p[U_0]$ is the required evenly covered neighbourhood of $p(x_0)$, as $p^{-1}[p[U_0]] = \oplus_{i=0}^n g_i[U_0]$, all of which are homeomorphic open sets, and $p$ is a homeomorphism between $g_i[U_0]$ and its image.
One way to get such a $U_0$ is to take pairwise disjoint neighbourhoods $V_i$ of the $x_i$ by Hausdorffness and take, as you propose, $U_0 = \bigcap_{i=0}^n g_i^{-1}[V_i]$. This is as required as $g_i[U_0] \subseteq V_i$ for all $i$, so the family of its images is pairwise disjoint.
The only requirements for a covering map are evenly covered neighbourhoods and being continuous and onto. The latter two are by construction (all quotient maps obey it) and the first I just proved. No local connectedness is required.