I was told to prove the following: If an action is free and satisfies that each point has a neighborhood $U$ satisfying $U \cap gU=\emptyset$ except for finitely many $g\in G$, and moreover the space is Hausdorff, then each point has a neighborhood such that $g_1\neq g_2\implies g_1U\cap g_2U=\emptyset$.
I'm pretty lost here - I don't have any intuition for why Hausdorfness is relevant either..
Let $x\in X$, consider a neigborhood $U$ of $x$ such that $V_x=\{g\in G:g(U)\cap U\neq \phi\}$ is finite. Write $V_x=\{y_1,...,y_n\}$ there exists $x\in U_i, y_i\in V_i$ such that $U_i\cap V_i$ is empty since $X$ is separated, set $W=U\cap_iU_i$, $W$ is an open subset containing $x$. Remark that if $g\in G$ and $g\neq I_d$ and $g(W)\cap W$ is not empty, then there exists $z\in W$ such that $g(z)\in W$, since $W\subset U$, $z\in V_x$, this is impossible since $V_x \cap W$ is empty since $W\cap V_i$ is empty.
let $g_1,g_2\in G, g_1\neq g_2$, consider $z\in g_1(W)\cap g_2(W)$. This is equivalent to saying that $z=g_1(w_1)=g_2(w_2)$ where $w_1,w_2\in W$, we deduce $w_1=g_1^{-1}g_2(w)$ this is a contradiction since $g_1^{-1}g_2(W)\cap W$ is empty.