How to prove the space of orbits is a Hausdorff space Let $M$ be a connected smooth n-dimensional manifold and $G$ a lie group acting smoothly on $M$.
for $x\in M$, the orbit $G\cdot x=\{g(x)\mid g\in G\} $ is a sub-manifold of $M$ and if the action is proper, namely
the inverse image of every compact subset of $M\times M$ under the map 
$$ G\times M\rightarrow M\times M: (g,p)\mapsto(p,g(p))$$
is compact.
Please  prove  the space of  orbits is a Hausdorff space.\
Thanks in advance.
 A: The crucial fact is that for every compact $K \subset M$, the set
$$G\cdot K = \{ g(x) : x \in K, g \in G\}$$
is closed. Suppose $y \in \overline{G\cdot K}$. Denote that map $(g,p) \mapsto (p,g(p))$ by $\varphi$. Then, for every compact neighbourhood $C$ of $y$, the set
$$B_C := \varphi^{-1}(K\times C)$$
is compact - since $K\times C$ is compact - and nonempty, since $C$ is a neighbourhood of $y$. The family $\mathcal{B} = \{ B_C : C \text{ compact neighbourhood of } y\}$ has the finite intersection property (since every finite intersection of compact neighbourhoods of $y$ is a compact neighbourhood of $y$), so
$$B_y := \bigcap_{C} B_C \neq \varnothing.$$
But
$$B_y = \bigcap_C \varphi^{-1}(K\times C) = \varphi^{-1}\left(\bigcap_C K\times C\right) = \varphi^{-1}(K\times \{y\}),$$
so there is an $x\in M$ and a $g \in G$ with $\varphi(g,x) = (x,g(x)) \in K \times \{y\}$, and that means $y \in G\cdot K$.
For $K = \{x\}$, that implies that the orbit $G\cdot x$ is closed.
For $x,y \in M$ with $G\cdot x \neq G\cdot y$, we have $G\cdot x \cap G\cdot y = \varnothing$. In particular, since $G\cdot y$ is closed, there is a compact neighbourhood $C$ of $x$ with $C \cap G\cdot y = \varnothing$. That implies $G\cdot C \cap G\cdot y = \varnothing$. $G\cdot C$ is closed, hence there is a  neighbourhood $U$ of $y$ with $U \cap G\cdot C = \varnothing$. Then also $G\cdot U \cap G\cdot C = \varnothing$. And that means that the images of $G\cdot C$ and $C\cdot U$ are disjoint neighbourhoods of $G\cdot x$ resp. $G\cdot y$ in $M/G$.
