Let $X$ be Haudorff, $G\subset \operatorname{Homeo}(X)$.Suppose that for any $x,y\in X\, \, \exists U,V \, \, \text{Open neigborhoods of $x$ and $y$ respectively.}$ such that $g\cdot U \cap V \neq \emptyset $ for at most finitely many $g\in G$. Show that $X{/}G$ is Hausdorff.
Proof
Consider $\pi:X \to X/G$ given by $x\to G_{x}$ where $G_{x}$ is the orbit of $x$.
Let $p,q\in X/G$ where $p\neq q$, then there exists $x,y\in X$ such that $\pi(x)=p$ and $\pi(y)=q$.
Consider $g_1,g_2,\cdots g_n\in G$ such that $g\cdot U \cap V \neq \emptyset$ for these $x,y\in X$. Since the space $X$ is Hausdorff there exists open neighborhoods $U_i,V_i\subset X$ such that $x\in U_i$ and $g_i\cdot y\in V_{i}$,$i\in I $.Where $I=\lbrace 1,2,\cdots n\rbrace $$
Now consider $$U=X\bigcap \left(\bigcap_{i\in I}U_i \right)$$ and $$V=X \bigcap \left( \bigcap_{i\in I} g_i^{-1}V_i \right)$$ .
We claim that $U\cap V=\emptyset$.And in fact show that $G_{U},G_{V}$ are disjoint orbits of $U$ and $V$ that contains $p,q$ and hence $X/G$ is Hausdorff.
From here I can´t show my claim.
Any advice of how I should start the proof was vey helpful I´m starting to learn this kind of concepts.