When is a quotient by closed equivalence relation Hausdorff

$\def\RR{\mathbb{R}}$There is no separation condition which will do the job. That's a vague statement, so here is a precise one: There is a subset of $\mathbb{R}^2$ which (equipped with the subspace topology) does not have condition $\dagger$.

Proof: Let $A$ and $B$ be two disjoint dense subsets of $\mathbb{R}$, neither of which contains $0$. (For example, $\mathbb{Q} +\sqrt{2}$ and $\mathbb{Q}+\sqrt{3}$.) Let $$X = (A \times \RR_{\geq 0}) \cup (B \times \RR_{\leq 0}) \cup (\{0\} \times \RR_{\neq 0}) \subset \RR^2.$$ Define $(x_1,y_1)$ and $(x_2, y_2)$ to be equivalent if $x_1=x_2$ and, in the case that $x_1=x_2=0$, that $y_1$ and $y_2$ have the same sign.

Verification that this is a closed equivalence relation: $X^2$ is a metric space, so we can check closure on sequence. Let suppose we have a sequence $(x_n, y_n) \sim (x'_n, y'_n)$ with $\lim_{n \to \infty} x_n=x$, $\lim_{n \to \infty} y_n=y$, $\lim_{n \to \infty} x'_n=x'$ and $\lim_{n \to \infty} y'_n=y'$. We must verify that $(x,y) \sim (x',y')$. First of all, we have $x_n = x'_n$, so $x=x'$ and, if $x=x' \neq 0$, we are done. If $x=x'=0$, we must verify that $y$ and $y'$ have the same sign. But $y_n$ and $y'_n$ weakly have the same sign for all $n$, so they can't approach limits with different signs.

Verification that $X/{\sim}$ is not Hausdorff: We claim that no pairs of open sets in $X/{\sim}$ separates the images of $(0,1)$ and $(0,-1)$. Suppose such open sets exist, and let $U$ and $V$ be their preimages in $X$. Then there is some $\delta$ such that $(A \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset U$ and $(B \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset B$. Then $U \cap \RR \times \{ 0 \}$ is an open set which contains $(-\delta, \delta)$. By the density of $B$, there must be a point of $B \cap (- \delta, \delta)$ in $U \cap \RR$, and then this gives an intersection between $U$ and $V$.

Theorem. Suppose that $X$ is a compact metrizable space and $R$ is a closed equivalence relation on $X$. Then $X/R$ is also Hausdorff (in fact, even metrizable).

Proof. I will need the following definition:

Definition. The equivalence relation $R$ on a topological space $X$ is usc (upper semicontinuous) if every $R$-equivalence class $[x]\subset X$ is compact and for every $R$-equivalence class $[x]\subset X$ and every (open) neighborhood $U$ of $[x]$ in $X$, there exists a neighborhood $V$ of $[x]$ in $X$ such that whenever an $R$-equivalence class $[y]\subset X$ has nonempty intersection with $V$, then $[y]\subset U$.

Here is how to think of this condition in the case when $(X,d)$ is a compact metric space: First of all, each equivalence class should be closed in $X$. Secondly, for every sequence $[y_n]$ of equivalence classes in $X$, if $d_{min}([x], [y_n])$ converges to zero, then $d_{max}([x],[y_n])$ also converges to zero.

Here $$ d_{min}([x],[y])=\min \{d(x',y'): x'\in [x], y'\in [y]\} $$ and $$ d_{max}([x],[y])=\max \{d(x',y'): x'\in [x], y'\in [y]\} $$
(the latter is also known as the Hausdorff distance between $[x]$ and $[y]$).

The relevant results are Propositions 1 and 2 on page 13 in

Robert J. Daverman, "Decompositions of manifolds." Reprint of the 1986 original. AMS Chelsea Publishing, Providence, RI, 2007.

which state that if $X$ is Hausdorff (resp. metrizable) and $R$ is usc, then $X/R$ is also Hausdorff (resp. metrizable).

[I can add a proof of Proposition 1 since it is an elementary exercise, but Proposition 2 is too long.]

Note that if $X$ is compact and $R$ is closed, then every $R$-equivalence class in $X$ is also closed, hence, compact. It remains to show:

Lemma. If $(X,d)$ is a compact metric space, then every closed equivalence relation $R$ on $X$ is usc.

Proof. Let $[x]$ be an $R$-equivalence class in $X$, and $U\subset X$ its neighborhood. I will consider a sequence $V_n$ of open metric neighborhoods of $[x]$: $$ V_n= \{z\in X: d(z, [x])< 1/n\}. $$ Here $d(z,A)=\inf \{d(z,a): a\in A\}$.

Suppose that for each $n$, there exists an $R$-equivalence class $[y_n]$ such that $y_n\in V_n$ and $[y_n]\cap X\setminus U \ne \emptyset$. Then we find $y'_n\in [y_n]$ such that $y'\in K=X\setminus U$; $K$ is a compact subset. Then, after passing to a subsequence, $(y_n)$ converges to some $y\in [x]$ while $(y'_n)$ converges to some $y'\in K$. Since $K\cap [x]=\emptyset$, it follows that $(y,y')\notin R$. At the same time, $\lim_n (y_n,y'_n)= (y,y')$ and $(y_n, y'_n)\in R$. This contradicts the assumption that $R$ is closed.

Thus, there exists $n$ such that $V=V_n$ has the property that whenever $[y]\cap V\ne \emptyset$, $[y]\subset U$. Hence, $R$ is usc. qed

Combining the lemma with Propositions 1 and 2, we conclude the proof of the theorem. qed