Proposition 3

If a topological group $G$ operates properly on a topological space $X$, then the orbit space $X/G$ is Hausdorff. If also $G$ is Hausdorff, then $X$ is Hausdorff.

Let $C \subset X \times X$ be the graph of the equivalence relation $R$ defined by $G$ on $X$; then $C$ is the image of $G \times X$ under the mapping $\theta : (s, x) \to (x, s.x)$. Since $\theta$ is proper, $C$ is closed in $X \times X$ (Chapter I, § 10, no. 1, Proposition 1). Since the relation $R$ is open ($§ 2$, no. 4, Lemma 2), it follows that $X/G$ is Hausdorff (Chapter I, § 8, no. 3, Proposition 8).

Now suppose that $G$ is Hausdorff. Then the mapping $x \to (e, x)$ of $X$ into $G \times X$ is a homeomorphism of $X$ onto a closed subspace of $G \times X$, and is therefore proper (Chapter I, § 10, no. 1, Proposition 2). If we compose this mapping with the mapping $(s, x) \to (x, s.x)$ of $G \times X$ into $X \times X$, which by hypothesis is proper, we obtain a proper mapping of $X$ into $X \times X$, namely the diagonal mapping $x \to (x, x)$. Hence the diagonal $\Delta$ of $X \times X$ is closed in $X$, and therefore $X$ is Hausdorff.