I am reading Emily Riehl's book Category Theory in Context. There is a chapter (Chapter 5) devoted on monads and their algebras. One of the exercise (Exercise 5.6.i) asks us to prove the fact that the forgetful functor $\mathsf{cHaus}\to\mathsf{Set}$ from the category of compact Hausdorff spaces to the category of sets is monadic, by using the fact that this functor is a conservative right adjoint and the following version of the monadicity theorem:
Let $U:\mathsf{D}\to\mathsf{C}$ be a conservative right adjoint functor. Suppose that $\mathsf{D}$ has coequalizers of $U$-split pairs and $U$ preserves them. Then $U$ is monadic.
(A $U$-split pair in $\mathsf{D}$ is a parallel pair whose image under $U$ has a split coequalizer.)
In view of the above theorem, we want to prove the following claim:
Claim: If $f,g:X\rightrightarrows Y$ is a parallel pair in $\mathsf{cHaus}$ which has a split coequalizer in $\mathsf{Set}$, and if $h:Y\to Z$ denotes the coequalizer of this pair in $\mathsf{Set}$, then $Z$ is a compact Hausdorff space with respect to the quotient topology induced by $h$.
One way to prove the above claim is to consider a topological space to be a pair of a set and a closure operator satisfying Kratowski's axioms*1. Since this proof is already given in the book, the author asks us instead to prove the above claim in the following way:
Hint: By the well-known result for compact Hausdorff spaces, it suffices to show that the equivalence relation on $Y$ induced by $h$ is closed, that is, the subset $R=\{(y_1,y_2)\in Y\times Y\mid h(y_1)=h(y_2)\}\subset Y\times Y$ is closed in $Y\times Y$. Use the split coequalizer diagram in $\mathsf{Set}$ to prove this.
If $s:Z\to Y$ and $t:Y\to X$ is a splitting for the fork $X\substack{ f\\ \rightrightarrows\\ g}Y\xrightarrow{h}Z$, so that $hs=1_Z,\,gt=1_Y,$ and $sh=ft$, and if one knows that at least one of $s$ or $t$ is continuous, then the above claim is not difficult to prove. Indeed, since $s$ is a (split) monomorphism in $\mathsf{Set}$, it is injective and thus we have $R=\{(y_1,y_2)\in Y\times Y\mid sh(y_1)=sh(y_2)\}=\{(y_1,y_2)\in Y\times Y\mid ft(y_1)=ft(y_2)\}$. If $s$ or $t$ is continuous, therefore, $sh=ft:Y\to Y$ will also be continuous, and since $Y$ is Hausdorff, it follows immediately that $R$ is closed in $Y\times Y$. But of course, it could happen that neither $s$ nor $t$ is continuous, and in this case I am at a loss. Can anyone help me prove the above claim in the way suggested in the hint? Any help is appreciated. Thanks in advance!
*1 Since split coequalizers are absolute coimits, the covariant power set functor $P:\mathsf{Set}\to\mathsf{Set}$ maps $h$ to the coequalizer of $Pf$ and $Pg$, and from this one easily sees that there is a topology on $Z$ which makes $h$ a continuous closed map. Since a surjective closed map is necessarily a quotient map, this is precisely the quotient topology induced by $h$. Because $h$ is closed, points of $Z$ are closed, and this, combined with the fact that $Y$ is normal, proves that $h$ is Hausdorff.