What are the epimorphisms in the category of Hausdorff spaces? It appears to be the case that the epimorphisms in $\text{Haus}$ are precisely the maps with dense image. This is claimed in various places, but a comment on my blog has made me doubt the source I got my proof from (Borceux).
Borceux's argument crucially uses the following result:

If $A \subset X$ is a closed subspace of a Hausdorff space $X$, then the quotient $X/A$ is Hausdorff. 

This appears to be false. As far as I can tell, if $X/A$ is Hausdorff, then $A$ and points in $X$ not in $A$ must be separated by open neighborhoods in $X$. But if this is true for every closed subspace $A$ of $X$, then $X$ is necessarily regular, and there are examples of Hausdorff spaces that aren't regular.
So: is it still true that the epimorphisms are precisely the maps with dense image? If so, what is a correct proof of this? 
 A: This should be a comment rather than an answer, but I don't have enough rep.  
HTop is actually the largest subcategory of Top closed under finite limits (as computed in Top) where all maps with dense image are epi:   
If $X$ is not Hausdorff then the equalizer of the projections $\pi_{1},\pi_{2}:X\times X\rightarrow X$, which is just the diagonal $\delta:X\rightarrow X\times X$, is not closed.  (Recall that a space is Hausdorff iff the diagonal is closed.)  
Let $C$ denote the closure of the diagonal in $X\times X$, let $d$ denote the factorization of $\delta$ through $C$, and let $p_{1}$ and $p_{2}$ denote the restrictions of $\pi_{1}$ and $\pi_{2}$ to $C$ respectively.  Then the image of $X$ is dense in $C$ and $p_{1}\circ d = p_{2} \circ d$, but $p_{1}\neq p_{2}$, so $d$ is not epi. 
This shows the fact Andy mentions in the comments, that equalizers are closed subspaces in HTop, is essential.
A: Yes, according to Herrlich & Strecker, Section 6.10(4). Here’s the argument:

If $A\overset{f}\longrightarrow B$ is an epimorphism, let $C$ be the disjoint topological union of two ‘copies’ of $B$ where the corresponding points of the closure of $f[A]$ have been identified, and let $h$ and $k$ be the two natural maps from $B$ to $C$. 

That’s as far as they actually write it out, but clearly the rest is that $h\circ f=k\circ f$, and $f$ is an epimorphism, so $h=k$, and $f[A]$ must therefore be dense in $B$.
Added: To see that $C$ is actually Hausdorff, let the copies of $B$ be $B_0=B\times\{0\}$ and $B_1=B\times\{1\}$, let $K=\operatorname{cl}f[A]$, and let $K_i=K\times\{i\}$ for $i\in\{0,1\}$. Finally, let $q:B_0\sqcup B_1\to C$ be the quotient map. Clearly $q(\langle x,i\rangle)$ and $q(\langle y,j\rangle)$ can be separated by disjoint open sets in $C$ whenever $x\ne y$, irrespective of $i$ and $j$. If $q(\langle x,0\rangle)\ne q(\langle x,1\rangle)$, then $x\in B\setminus K$, an open subset of $B$, so $q[B_0\setminus K_0]$ and $q[B_1\setminus K_1]$ are disjoint open nbhds of $q(\langle x,0\rangle)$ and $q(\langle x,1\rangle)$.
A: In Topology and Groupoids, p. 128, it is proved that an adjunction space $B \; _f\sqcup X$ is Hausdorff if (a) $B$ and $X$ are Hausdorff, (b) each $x \in X \backslash A$ has a neighbourhood closed in $X$ and not meeting $A$, and (c) $A$ is a neighbourhood retract of $X$. 
I do not know if these conditions can be weakened for the case of $X/A$. 
I should say this is really a comment on the comments rather than an answer to the question! 
