= Solution
A <monad> on $\mathcal C$ consists of an <endofunctor> $T$ and <natural transformations> $\eta:1\to T$, $\mu:T^2\to T$ satisfying $\mu\,T\eta=\mu\,\eta T=1_T$ and $\mu\,T\mu=\mu\,\mu T$. Its <Eilenberg-Moore category> has objects $(X,a)$ with $a:TX\to X$, $a\eta_X=1_X$ and $a\mu_X=aTa$. A morphism $f:(X,a)\to(Y,b)$ satisfies $fa=bTf$. An <adjunction> $F\dashv G$, with unit $\eta$ and counit $\varepsilon$, induces $T=GF$, $\mu=G\varepsilon F$. It is a <monadic adjunction> if the <Eilenberg-Moore comparison functor> $K(D)=(GD,G\varepsilon_D)$ is an equivalence with the <Eilenberg-Moore category>.
A fork $A\overset{p,q}{\rightrightarrows}B\overset{r}{\to}C$ is a <split coequalizer> if $rp=rq$ and there are $s:C\to B$, $t:B\to A$ with $rs=1$, $pt=1$ and $qt=sr$. It is indeed a <coequalizer>: if $hp=hq$, then $h=hpt=hqt=hsr$, so $h$ factors uniquely through $r$, which is split epic. Every <functor> preserves these splitting identities, making such <coequalizers> absolute.
The <precise monadicity theorem> states: for an <adjunction> $F:\mathcal C\rightleftarrows\mathcal D:G$, the comparison is an equivalence if and only if $G$ reflects <isomorphisms> and every pair in $\mathcal D$ whose $G$-image admits a <split coequalizer> has a <coequalizer> which $G$ preserves. Equivalently, the right adjoint creates these split-underlying <coequalizers>, with creation understood up to the canonical <isomorphism> of the prescribed underlying <coequalizer>. Here is a proof.
First consider the forgetful <functor> from <monad algebras>. It reflects <isomorphisms>: the inverse of an invertible algebra morphism also obeys the algebra-morphism identity, by multiplying that identity by the inverses. For an algebra pair with a split underlying <coequalizer> $r:B\to C$, all powers of $T$ preserve its <coequalizer>. If $b:TB\to B$ is the algebra structure on $B$, the map $rb$ coequalizes $Tp,Tq$, so there is a unique $c:TC\to C$ with
$$
cTr=rb.
$$
The unit law follows by composing with the <epimorphism> $r$: $c\eta_Cr=cTr\eta_B=rb\eta_B=r$. For associativity, naturality and the algebra law give
$$
c\mu_C T^2r=cTr\mu_B=rb\mu_B=rbTb=cT(rb)=cTcT^2r.
$$
Cancel the <epimorphism> $T^2r$ to obtain $c\mu_C=cTc$. Thus $(C,c)$ is an algebra and $r$ is an algebra morphism. Any algebra morphism coequalizing the pair factors through $r$ on underlying objects, and its factor is an algebra morphism by cancellation of the <epimorphism> $Tr$. This creates the <coequalizer>. Transporting these facts through an equivalence proves necessity.
For sufficiency, let $(X,a)$ be a $T$-algebra. In $\mathcal D$ consider the free pair
$$
FTX\ \underset{Fa}{\overset{\varepsilon_{FX}}{\rightrightarrows}}\ FX.
$$
Its underlying pair is $\mu_X,Ta:T^2X\rightrightarrows TX$, with <split coequalizer> $a:TX\to X$: the splittings are $s=\eta_X$ and $t=\eta_{TX}$, using the algebra unit law, the <monad> unit law and naturality of $\eta$. By hypothesis its <coequalizer> $q:FX\to D$ exists and is preserved. Identify $GD$ with $X$ by the unique <coequalizer> <isomorphism>, so $Gq=a$. Naturality of the counit makes $Kq$ an algebra morphism, hence the comparison structure $c$ on $GD$ satisfies $cTa=a\mu_X=aTa$. The map $Ta$ is split epic because $a\eta_X=1$, so $c=a$. Every algebra is therefore isomorphic to a comparison algebra.
For an actual object $D$, its counit $\varepsilon_D:FGD\to D$ is a <coequalizer> of the analogous free pair. It coequalizes that pair by counit naturality. Compare it with the <coequalizer> furnished by the hypothesis; applying $G$ gives an <isomorphism>, since both underlying arrows coequalize the same split pair. Reflection of <isomorphisms> makes the comparison map an <isomorphism>. This proves the counit's asserted <coequalizer> property.
Let $f:KD\to KE$ be an algebra morphism. Its transpose is $h_0=\varepsilon_EFf:FGD\to E$. The two composites of $h_0$ with the free pair have adjunct transposes $G\varepsilon_E\,Tf$ and $fG\varepsilon_D$, equal because $f$ is an algebra morphism. Consequently $h_0$ factors uniquely through $\varepsilon_D$ as $h:D\to E$. Applying $G$ gives $(Gh)G\varepsilon_D=fG\varepsilon_D$; the last factor is split epic, so $Gh=f$. Conversely any lift of $f$ must satisfy $h\varepsilon_D=\varepsilon_EFf$ by naturality and hence is unique. Thus $K$ is full and faithful, and the earlier construction makes it essentially surjective. The comparison is an equivalence, completing the theorem's proof.
Now take $G:\mathbf{CompHaus}\to\mathbf{Set}$. Its left adjoint exists by the permitted topological result; it can be realized as the <Stone-Čech compactification> of the discrete input set. A bijective <continuous map> from a <compact space> to a <Hausdorff space> is a <homeomorphism>, so $G$ reflects <isomorphisms>.
For continuous $f,g:X\rightrightarrows Y$ with a split underlying fork $q,s,t$, let $R$ be the relation of having the same $q$-image. We claim
$$
R=\{(f(x),f(x')):g(x)=g(x')\}.
$$
For the forward inclusion choose $x=t(y)$, $x'=t(y')$: $ft=1$ and $gt=sq$ supply the equality. For the reverse inclusion use $qf=qg$. The subset $\{(x,x'):g(x)=g(x')\}\subset X^2$ is closed, because $Y$ is Hausdorff, and compact. Its continuous image $R\subset Y^2$ is compact and hence closed. A closed <equivalence relation> on a <compact Hausdorff space> has a compact Hausdorff quotient; its quotient map is a <coequalizer> here, since it identifies exactly the generated relation. Its underlying set <coequalizer> is the specified one up to the unique <isomorphism>. Thus $G$ preserves the needed <coequalizers>, and
$$
\boxed{\mathbf{CompHaus}\rightleftarrows\mathbf{Set}\text{ is a monadic adjunction}.}
$$
The closedness argument is the <closed relation generated by a set-split pair of compact Hausdorff maps> lemma; the splitting maps themselves need not be continuous.
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