# If the orbit space is Hausdorff, must the action be proper?

Let $G$ be a Lie group acting continuously on a manifold $M$. It is well known that if the action is proper, then the quotient space $M/G$ is Hausdorff.

Does the converse hold? If $M/G$ is Hausdorff, must the action be proper?

No, let $G$ be a non locally compact Lie group (infinite dimensional) the left action of $G$ on $G$ is not proper, for example, take any neighborhood $U$ of the identity $1$, the set $\{g\in G, g(U)\cap U\neq \phi\}$ is infinite. but the quotient of $G$ by $G$ is a point.
An interesting question is to know if your question is true when $G$ is a discrete group acting on a manifold $M$.
This is a silly counterexample but it's a counterexample. Let $M=\{0\}$ and let $G=\mathbb{R}$ act on $M$ by multiplication. Then $M/G$ is still only one point but the action is not proper, since the preimage of $(0,0) \in M\times M$ is $G\times M$.
This is a instant of a more general phenomena, namely that if $M$ is compact and $G$ is not then the action is not proper. If it happens (as in this case) that the action is transitive your quotient will be one point and thus Hausdorff.