Quotient of a locally compact Hausdorff space by a proper action is Hausdorff

Question

I am trying to prove the following:

Let $G$ be a topological group acting properly on a Hausdorff locally compact space $X$, i.e. preimages of compacts sets by the map $$G\times X\to X\times X$$ $$(g,x)\mapsto (gx,x)$$ are compacts. Then the quotient $X/G$ is Hausdorff.

Take $x,x'\in X$ such that $Gx\not=Gx'$.

1. By using that X is Hausdorff locally compact and property of the action, I found $V_1,V_2$ disjoint open set with compact closure and $H\subset G$ compact, such that $x\in V_1,y'\in V_2$ and $$(G-H)\overline{V_1}\cap \overline{V_2}=\varnothing.$$
2. So now we need to work on $H$. Since $x'\in (Hx)^c\cap V_2$ open, there exists $x'A_2\subset V_2$ such that $$A_2\cap Hx=\varnothing.$$ It implies that $x\in (H^{-1}A_2)^c\cap V_1$.
3. Now, I'd like to find as before $x\in A_1\subset V_1$ open such that $$A_1\cap H^{-1}A_2=\varnothing,$$ which would be enough to conclude. But $H^{-1}A_2$ is not closed like $Hx$, so I'm stuck...

Are my first ideas correct ? Is there a way to exit this dead end ?

Answer 1

So using that locally compact spaces are compactly generated, it is easy to show that a proper map $f: A\to B$ with $B$ locally compact is closed.

Therefore, $f: G\times X\to X\times X$ as defined above is closed. But the quotient map $\pi: X\to X/G$ is closed, so $$\tilde f=\pi\circ f: G\times X\to X/G\times X/G$$ is also closed. Note that $\tilde f(g,x)=([gx],[x])=([x],[x])$, so the diagonal $$\Delta(X/G)=\tilde f(G\times X)$$ is closed, i.e. $X/G$ is Hausdorff.

Is that correct ?