Let $X$ be a Hausdorff topological space and $G$ a group of homeomorphisms from $X$ to $X$. Suppose that $G$ acts on $X$ discretely, that is, for any $a,b$ in $X$, there exists a neighbourhood $U_x$ of $x$ which contains finitely many elements of the orbit of $a$. Moreover suppose that $G$ acts freely on $X$, that is, $g(x)\neq x$ for any $x\in X$ and any $g\in G\setminus\{id\}$. Show that $X/G$ (equipped with the quotient topology) is Hausdorff.
I tried as follows: let $Ga$ and $Gb$ be two different orbits of $a,b$ respectively, then neighbourhoods of these two orbits (in $G/X$) have the form $\pi(U), \pi(V)$ where $\pi:X\to X/G$ is the quotient map and $U,V$ are neighbourhoods of $a,b$ respectively. We need to find $U,V$ so that $\pi(U)\cap \pi(V)=\emptyset$. The latter is equivalent to the fact that $Orb(x)\cap Orb(y)=\emptyset$ for any $x\in U,y\in V$. This means that we need to find $U,V$ so that $g(U)\cap V=\emptyset$ for any $g\in G$. Since $G$ acts discretely on $X$, for each $x\in U$ we can find a small neighbourhood $V_x$ of $b$ so that $Orb(x)\cap V_x$ is finite. Then as $X$ is Hausdorff, one can choose $V_x$ such that $Orb(x)\cap V_x=\emptyset$. However when $x$ varies in $U$ the intersection of corresponding $V_x$ is not necessary a neighbourhood of $b$ and I get stuck here.
I guess the condition that the action is free plays some role, but I do not know how to use it. I hope someone can help me. Thanks a lot!