Is compactness (without Hausdorff) enough to get a closed quotient map?

Question

This is some attempt to generalise the classical result for a compact Hausdorff space $D$, if $E\subseteq D\times D$ is closed, then the quotient map $p:D\to D/{E}$ is a closed equivalence relation. Below $D$ is not Hausdorff anymore (but (quasi)-compact and $D\times D$ is considered with topology that is finer than product topology in terms of closed sets).

Let $D$ be a Noetherian topological space. Let $E\subseteq D\times D$ be a closed equivalence relation. Note that the topology on $D\times D$ is not necessarily the product topology, but it is a Noetherian topology which is finer than the product topology with respect to closed sets (e.g. as for algebraic varieties, I am aware that this is a rather vague statement. I am thinking on it to clarify further).

Consider $T := D/{E}$ with the natural quotient topology, and let $p:D\to T$ be the quotient map. Is it true that $p$ is a closed function, i.e. is the image of a closed set under $p$ closed?

Note that $D$ is not necessarily Hausdorff, but it is compact (or quasi-compact in algebraic geometry terminology) in the sense that every open cover has a finite subcover.

Answer 1

A classical theorem by Kuratowski: if $K$ is compact (quasi-compact in Bourbaki parlance, so no separation axioms) and $X$ is any space, $\pi_X: X \times K \to X$ is closed (where $X \times K$ has the product topology and $\pi_X$ is the projection onto the first coordinate), which is often formulated as "a projection along a compact space is closed". This property in fact characterises compactness : if $K$ is a space such that for any space $X$ this $\pi_X$ is closed, then $K$ is compact.

Now:

If $X$ is compact and $R \subseteq X \times X$ is a closed equivalence relation (where $X \times X$ has the product topology) the natural map $q: X \to X/{R}$ is a closed map, where $X/{R}$ has the quotient topology wrt $q$.

Proof Let $F$ be closed in $X$. Then a moment's thought shows that (letting $\pi_1: X \times X \to X$ be the projection onto the first coordinate):

$$q^{-1}[q[F]] = \pi_1[((X \times F) \cap R]$$

and so $q^{-1}[F]$ is closed in $X$ as $\pi_1$ is a closed map by the Kuratowski theorem and $(X \times F) \cap R$ is closed in $X \times X$ as $R$ is a closed set in the product. So $q[F]$ is closed in $X/{R}$ as the latter set has the quotient topology wrt $q$. So $q$ is closed. QED.