I can't find a counterexample to this sentence:
Let $X$ be a $T2$ topological space, and $G \subset Hom(X)$ be a group of homeomorphisms which acts properly discontinuously on X (i.e. $\forall x \in X \ \ \exists U$ neighborhood of $x$ such that $\forall g \in G\setminus \{id\} \ \ g(U) \cap U=\varnothing).$ Then the quotient $X/G$ is $T2.$
The following theorem gives me some hints about what to avoid while making the counterexample :
Let $A \subset X$ (with $X$ Hausdorff) be an open set such that: $A$ intersects every orbit of $G,$ and $A \cap g(A) \neq \varnothing$ for a finite quantity of $g \in G.$ Then $X/G$ is $T2.$
Clearly $G$ cannot be finite, otherwise the quotient is automatically $T2,$ taking $A=X.$ Moreover, this condition is somehow similar to the hypothesis of properly discontinuous action, and I can't find an example where they don't "coincide". Do you have some hints? Thanks to everybody.
This contradicts the theorem you posted, so I'm not sure if our definitions of properly discontinuous actions are the same. Anyhow,
$\mathbb{Z}/2$ acts on the disjoint union of $\mathbb{R} \times \{a\}$ and $\mathbb{R} \times \{b\}$ by $g \cdot (0, a) = (0, a), g \cdot (0, b) = (0, b)$, and otherwise $g \cdot (t, a) = (t, b)$ and vice-versa. This is a properly discontinuous action (the group is finite). Furthermore, the original space is certainly Hausdorff in the disjoint union topology; the resulting space is the line with two origins which is not Hausdorff at the origins.