Is it possible to show that every coequalizer in the category of Hausdorff spaces is a quotient map directly from the universal property of a coequalizer and without use of the set-theoretical construction of the standard-coequalizer?
More precisely, is there a replacement for $h_U$ in the following proof sketch such that $h_U$s codomain is a Hausdorff space?
All maps hereinafter shall be continuous functions, that is TOP-arrows.
Problem: Let $e:B\to E$ be a coequalizer for the arrows $f_1,f_2:A\to B$. $e$ is a quotient map, i.e. $\forall U \subseteq E: e^{-1}(U) \text{ open in } B\leftrightarrow U \text{ open in } E$
Proof: Let $U\subseteq E$ such that $e^{-1}(U)$ is open in $B$. I aim for $U$ to be open in $E$.
To that end I define a map $h_U:B\to D$, where $D$ is $U$ with a new point $0$ adjoint to it in an indiscrete fashion, that is the topology of $D=\mathcal{O}_E(U) \cup \{U,U\cup\{0\}\}$, where $\mathcal{O}_E(U)$ is the subspace topology of $U$ regarded as a subspace of $E$.
I want $h_U$ to be a sort of indicator function for $e^{-1}(U)$ so I set $h_U(x) = \begin{cases}e(x) & x\in e^{-1}(U) \\ 0 & \text{otherwise}\end{cases}$.
Now according to my calculations $h_U\circ f_1 = h_U\circ f_2$ and the universal property of the coequalizer $e$ yields an arrow $j_U:E\to D$ with $h_U = j_U\circ e$ which renders $U=j_U^{-1}(U)$ open.
Background: I want to show that every coequalizer is a quotient map in the context of a set theory which has a notion of open sets, but is not powerful enough to grant me standard-coequalizers.