I feel this ought to be a very simple question, but I seem to be asking it at a greater level of generality than people usually do:
Is the orbit space $|X|$ of every proper topological groupoid $G \rightrightarrows X$ Hausdorff if $X$ itself is Hausdorff?
I put emphasis on the word "every" in that question, because in all the references I can find, the question is not answered without placing at least one or two extra assumptions on the groupoid, and what I am wondering is whether those extra assumptions are truly necessary.
Let me first clarify my definitions, and then point out where I see some subtlety. By a topological groupoid $G \rightrightarrows X$, I mean a small category whose object set $X$ and morphism set $G$ are both topological spaces, the latter containing only isomorphisms, such that the standard structure maps
- source and target $s,t : G \to X$, which send each morphism $x \stackrel{g}{\to} y$ to $x$ and $y$ respectively;
- composition $c : \{ (g,h) \in G \times G \ |\ s(g) = t(h) \} \to G : (g,h) \mapsto g \circ h$;
- unit $u : X \to G$, which sends each object $x$ to the identity morphism $x \to x$;
- inversion $i : G \to G : g \mapsto g^{-1}$
are all continuous. The orbit space $|X| = X /{\sim}$ is the quotient defined by calling $x,y \in X$ equivalent whenever there exists a morphism between them, and I will write the equivalence class represented by $x \in X$ as $|x| \in |X|$. Two other conditions are especially relevant to my question: we say that $G \rightrightarrows X$ is
- proper if $(s,t) : G \to X \times X$ is a proper map;
- open if $s : G \to X$ and $t : G \to X$ are open maps (in fact one can assume this for only one of them, and it then follows for the other).
So for example, the action groupoid of a continuous action $H \times X \to X$ of a topological group $H$ on a Hausdorff space $X$ can be defined to have morphism space $G := \{ (x,h,y) \in X \times H \times X \ |\ y = hx \}$ with source map $s(x,h,y) = x$, target $t(x,h,y) = y$ and more-or-less obvious definitions for the rest of the structure maps. An action groupoid is always open, and it is proper if and only if the group action is proper. The orbit space $|X|$ is in this case the quotient $X/H$, which is indeed Hausdorff if the action is proper.
Here's where I find things getting tricky: as observed in this question, not every topological groupoid is open, and for those that aren't, the quotient projection $X \to |X|$ need not be an open map. I do know how to prove the following:
Theorem: If $X$ is Hausdorff and $G \rightrightarrows X$ is open and proper, then $|X|$ is also Hausdorff.
In the slightly simpler setting where $X$ and $G$ are both assumed first countable, the openness of the quotient projection implies that $|X|$ is also first countable, and a quick proof then goes as follows: suppose $|x_n| \in |X|$ is a sequence converging to two elements $|x|,|y| \in |X|$. Appealing again to the openness of the quotient projection, it follows that two sets of representatives $x_n,y_n \in X$ for this sequence can be chosen such that $x_n \to x$ and $y_n \to y$ in $X$, and the fact that $|x_n| = |y_n| \in |X|$ means there also exists a sequence of morphisms $g_n \in G$ with $s(g_n)=x_n$ and $t(g_n)=y_n$. Properness then gives $g_n$ a subsequence that converges to some $g \in G$, and the continuity of $(s,t)$ then implies $x_n \to s(g)$ and $y_n \to t(g)$. Since $X$ is Hausdorff and limits of sequences in $X$ are therefore unique, it follows that $s(g)=x$ and $t(g)=y$, so $g$ is a morphism $x \to y$, proving $|x|=|y|$, i.e. limits of sequences in $|X|$ are also unique.
If $X$ and $Y$ are not assumed first countable, then the argument above still works if one replaces sequences across the board with nets; one just needs to have the right notion of what it means for a continuous map $f : X \to Y$ between arbitrary topological spaces to be proper, i.e. it should mean that $f$ is a closed map for which $f^{-1}(y)$ is compact for all $y \in Y$. (If $Y$ is Hausdorff, that is equivalent to the condition that any net in $X$ whose image under $f$ converges must have a convergent subnet. If additionally $X$ and $Y$ are first countable, then it is equivalent to the usual "$f^{-1}(\text{compact}) = \text{compact}$," but in general the latter is a weaker condition.)
But if the quotient map $X \to |X|$ is not open, then I don't immediately see what to do with the assumption that $|x_n| \in |X|$ converges to both $|x|$ and $|y|$, because it is much less straightforward to characterize convergence in the quotient in terms of convergence in $X$. My instinct is rather to look for a counterexample, but I'm not sure how to cook up a groupoid that would be strange enough.
A couple of final remarks about the literature I've seen: the vast majority of people who think about proper groupoids are interested specifically in Lie groupoids, which are always open since the source and target maps are submersions, so no problem there. There is also no problem with étale topological groupoids in general, which are likewise automatically open. Otherwise, it seems that some version of my question is addressed in this paper by Jean-Louis Tu, but it focuses on groupoids that are locally compact, and that's an assumption that I especially do not want to make. (Most groupoids I'm interested in that I would call "topological" rather than "Lie" groupoids are infinite dimensional, so definitely not locally compact.)
Revised addendum: Many thanks to Moishe Kohan for pointing out in the comments why what I said in the crossed-out text below could not be completely correct, which made me notice a few places where I had sneakily assumed $X$ was Hausdorff without saying so. I have now revised the text of the question accordingly.
Addendum: You may have noticed that I didn't assume $X$ or $G$ were Hausdorff in the theorem above. I would have intuitively expected that to be necessary, but it doesn't seem to be, and as a sanity check, it is an amusing exercise to show that if you take any space $X$ and turn it into a topological groupoid in the trivial way with only the identity morphisms, then that groupoid cannot be proper unless $X$ is Hausdorff (and vice versa)!
EDIT: It took me about 12 hours to realize that the "counterexample" I described below is not a counterexample, but is in fact an open and proper groupoid with a perfectly reasonable Hausdorff orbit space. (The topology on the orbit space is not what I claimed below: the two points that I thought always had intersecting neighborhoods are actually discrete points, and sequences $(x_k,y_k) \in X$ with $y_k \to \pm 1$ do not converge after projecting to the orbit space.) So, my question remains unanswered... instead of deleting the wrong answer, I will leave it here as a demonstration of what not to do.
OK, I think I can now answer the question myself, so I'll record the answer here.
There are counterexamples: in other words, there are proper topological groupoids $G \rightrightarrows X$ for which $X$ and $G$ are both Hausdorff but the orbit space $|X|$ is not. They are, of course, not open.
I'll try to describe a concrete counterexample.
As was pointed out in an answer to the question on open groupoids that I previously mentioned, any equivalence relation $\sim$ on a topological space $X$ gives rise to a topological groupoid $G \rightrightarrows X$ for which $|X| = X/{\sim}$: one defines it so that there is at most one morphism $x \to y$ for any two $x,y \in X$, and it exists if and only if $x \sim y$. To make the topology of the morphism space completely transparent, we can define it explicitly as $$ G := \left\{ (x,y) \in X \times X \ \big|\ x \sim y \right\} $$ with source and target maps $s(x,y) = x$ and $t(x,y) = y$. Assuming $X$ is Hausdorff and first countable, the groupoid $G \rightrightarrows X$ is now proper if and only if $G$ is a closed subset of $X \times X$.
If we want a concrete example that completely breaks the proof I described that $|X|$ is Hausdorff, it needs to have the property that there exist convergent sequences $|x_n| \to |x|$ in $|X|$ for which the equivalence classes cannot be represented by any convergent sequence $x_n \to x$ in $X$. Here's an example that has that property and also a non-Hausdorff quotient.
Choose any decreasing continuous function $f : {\mathbb R} \to (1,\infty)$ such that $f(x) \to 1$ as $x \to \infty$, and define $X_+$ to be the following subset of the upper half-plane in ${\mathbb R}^2$: $$ X_+ := ({\mathbb R} \times \{1\}) \cup \left\{ (x,y) \in {\mathbb R}^2 \ \big|\ y > f(x) \right\}. $$ Now define $X \subset {\mathbb R}^2$ to be the union of $X_+$ with its reflection across the $x$-axis, and impose on $X$ the smallest equivalence relation such that $$ (x,y) \sim (x',y) \quad \text{ for all } \quad x,x',y, $$ and $$ (x,y) \sim (x,-y) \text{ for all } x,y \text{ with } y > 1. $$ Thus the equivalence class of each point $(x,y) \in X$ is determined by its $y$-coordinate, and points whose $y$-coordinates differ by a sign are also equivalent, with the important exception of the two lines $y = \pm 1$, which are distinct equivalence classes. Topologically, $X/{\sim}$ is homeomorphic to something that looks like $[0,\infty)$ except with the boundary point $0$ replaced by two distinct points whose neighborhoods always intersect each other. (In other words, $X/{\sim}$ would be a 1-manifold with boundary if we didn't require those to be Hausdorff.)
The groupoid $G \rightrightarrows X$ constructed out of this equivalence relation is nonetheless proper, because the condition $y > f(x)$ in the definition of $X_+$ means that no convergent sequence $(x_k,y_k) \in X$ can have equivalence classes in $|X|$ converging to $\{ y=\pm 1\}$; sending $y_k$ to $\pm 1$ forces $x_k \to \infty$. As a result, the question of whether $G$ is a closed subset of $X \times X$ simply doesn't see the fact that the topology of $|X|$ behaves badly at those two points.
If I've overlooked something crucial in this answer, then it wouldn't be the first time, so all comments are welcome!
(Addendum: As mentioned in the edit at the top of this answer, I did indeed overlook something crucial. My instict still says one ought to be able to cook up a counterexample in the way I had in mind, but this one isn't strange enough.)