I'm trying to understand the final step in the following argument showing the epimorphisms in $\textbf{Haus}$ have dense image. Let $C$ be a topological space. For $x, y \in C$, let $x \sim y$ if and only if for any continuous map $g : C \to D$ with $D$ Hausdorff, we have $g(x) = g(y).$ We call $H(C) = C/\sim$ the Hausdorff quotient of $C$.
Let $f: A \to B$ be an epimorphism in $\textbf{Haus}$. Let $E = \overline{f(A)}$, the closure of the image $f(A)$ in $B$. Let $q: B \to B/E$ be the quotient map. Let $r: B/E \to H(B/E)$ take $B/E$ to its Hausdorff qoutient. Then $r \circ q \circ f$ is constant, since $q$ identifies all points in the image of $f$ and $r$ must respect this equivalence class. Let $c: B \to H(B/E)$ be the constant function sending $b \in B$ to the value $r \circ q \circ f(a)$. Then $r \circ q \circ f = c \circ f$ and since $f$ is an epimorphism, $r \circ q$ must also be constant. Thus either $q$ is constant or $r$ is constant, since $q$ is a surjection. If $q$ is constant, then $B = E$. If $r$ is constant and $B \neq E$, suppose $x \in B\backslash E$. Then there exists an open set $O$ with $x \in O \subseteq B\backslash E$. For $y, z$ in $O$ we have $q(y) = q(z)$ if and only if $y = z$. Thus $q^{-1}(q(O)) = O$, since if $x' \in q^{-1}(q(O))$, we have $q(x') = q(x)$ for some $x \in O$ and by the definition of $q$, either $x,x' \in E$, which cannot hold since $x \in O$, or $x = x'$ and $x' \in O$. As $B/E$ has the quotient topology with respect to $q$ and $O$ is a saturated open set, $q(O)$ is open in $B/E$. We show that $q(O)$ is Hausdorff. Let $a',b' \in q(O)$ be distinct points. Then there exist distinct $a,b \in O$ with $q(a) = a'$ and $q(b) = b'.$ Since $B$ is Hausdorff $O$ is Hausdorff and we can find disjoint open subsets $U,V \subseteq O$ with $a \in U$ and $b \in V$. Then we have that $U,V$ are saturated open with respect to $q$. Since $q$ is injective on $O$, $q(U) \bigcap q(V) = \emptyset$. Therefore $q(U)$ and $q(V)$ are open and disjoint in $B/E$ with $a' \in q(U)$ and $b' \in q(V)$, making $q(O)$ an open Hausdorff subset of $B/E$. Therefore, $r \circ q(O) \neq r \circ q (E)$, contradicting the assumption that $r$ is constant.
I do not understand this last conclusion.