Separating an orbit and a point by neighborhoods Let $G$ be a group acting on the Hausdorff topological space $X$ discontinuously (meaning each orbit does not contain any limit point in $X$). Let $x,y\in X$ be two points that are not $G$ equivalent, meaning there is no $g\in G$ such that $g.x=y$. I wonder how to show the following:

There exists open neighborhoods $U$ of $x$  and $V$ of $y$ such that $gU\cap V=\varnothing$ for all $g\in G$. 

It is easy to see that we can find an open neighborhood $V$ of $y$ such that $G.x\cap V=\varnothing$. But I don't know how to continue.
Source: Iwaniec topics in classical automorphic forms, page 29.

 A: I think the stuff from the picture is bad.
As I understood the definition, $\Gamma$ acts on $X$ transitively. This means that an orbit $\Gamma x=X$ for any point $x\in X$, see, for instance, Wikipedia. So each two points of $X$ are $\Gamma$-equivalent and all formal claims about not $\Gamma$-equivalent points of $X$ trivially true. :-)  
The stability groups of points need not to be finite. For instance, let $X$ be any infinite discrete space and $\Gamma$ be a group of all bijections of the set $X$. Then $\Gamma$ acts on $X$ transitively and discontinuously, but all stability groups $\Gamma_x$, $x\in X$  are infinite.
If $X$ has no limit points in $X$ then each point of $X$ is isolated. This makes the claim “for any $x\in X$ there exists a small neighborhood $U\dots$” also trivial. 
Trying to obtain non-trivial facts from the definitions, we can relax transitivity condition in the definition of a discontinuous action (as you did).

There exists open neighborhoods $U$ of $x$  and $V$ of $y$ such that $gU\cap V=\varnothing$ for all $g\in G$.

But this is not necessarily true. For a counterexample for the claim it suffices in example provided by Moishe Kohan put $x=(1,0)$ and $y=(0,1)$. Indeed, for each neighborhood $U$ of $x$ there exists $\varepsilon>0$ such that $A=\{1\}\times [0,\varepsilon)\subset U$. It is easy to see that $y\in\overline{GA}$.
