A monomorphism in the category of compact Hausdorff spaces is regular Let $f \colon X \rightarrow Y$ be a monomorphism of compact Hausdorff spaces. This is just a continuous injection. I am trying to show that $f$ is regular, i.e. it is an equaliser.
My first thought was to consider the constant map $f(X) \rightarrow [0,1]$ which maps everything to $0$. By Urysohn's Lemma, there exists a morphism $g \colon Y \rightarrow [0,1]$ which extends the other map. $f$ would be the coequaliser of $g$ and the constant zero map if we could write
$$
f(X) = \{ y \in Y \mid g(y) = 0\}
$$
But unfortunately I don't think we can guarantee that $g(y) \not=0$ for $y$ outside of $f(X)$. 
 A: First note that a monomorphism $f:X\to Y$ must be injective: If $f(x)=f(y)$, then the compositions of $f$ with the maps $(*\to X)$ sending the single point to $x$ and $y$, respectively, are equal, thus these maps are equal, hence $x=y$.
So let $f:X\to Y$ be a monomorphism in $\mathbf{CHaus}$. Since a map from a compact space to a Hausdorff is perfect (meaning that it's closed and all fibers are compact), all maps in $\mathbf{CHaus}$ are perfect maps. Now consider the quotient map $q:Y\to Y/f(X)$, collapsing the image of $f$ to a single point. This space is clearly compact, but it's also Hausdorff, as is every quotient of a Hausdorff regular space by a closed subset (alternatively you could use that perfect surjective maps preserve the Hausdorff property). Now if $c:Y\to Y/f(X)$ is the constant map to $\{f(X)\}$, then $q$ and $c$ agree precisely on $f(X)$. If $g:Z\to Y$ is any map such that $qg=cg$, then its image $g(Z)$ must be a subset of $f(X)$. Since $f$ is injective, $g$ factors uniquely as set maps $fh:Z\to X\to Y$. Finally, $h$ is continuous since $f$ is a (closed) embedding.
A: Your idea to use Urysohn's Lemma can be made to work, in the following way.
First note that $f(X)$ being the continuous image of a compact space is compact  and thus closed (because $Y$ is Hausdorff).  In fact $f(C)$ is closed in $Y$ for every closed subset $C$ of $X$ for the same reason, a fact that will come in useful later on.
Now, since every compact Hausdorff space is normal, we can find, using Urysohn's Lemma, for every $y\in Y\backslash f(X)$ a continuous map $g_y\colon Y\to [0,1]$ with $g_y(y)=1$ and $g_y(f(x))=0$ for all $x\in X$. 
We combine these $g_y$'s into one continuous map $g\colon Y\to [0,1]^{Y\backslash f(X)}$ by setting $g(z)(y)=g_y(z)$. Note that $[0,1]^{Y\backslash f(X)}$ is a compact by Tychonoff's theorem.
Let $\nu  \colon Y\to [0,1]^{Y/f(X)}$ be the continuous map given by $\nu (z)(y)=0$.
We claim that $f\colon X \to Y$ is the equalizer of $\nu,g \colon Y\to [0,1]^{Y\backslash f(X)}$. 
To begin, we have $\nu\circ f = g\circ f$, because $\nu(f(x))(y)=0=g_y(f(x))$ for all $x\in X$ and $y\in Y\backslash f(X)$.
Let $Z$ be a compact Hausdorff space, and let $h\colon Z\to Y$ be a continuous map with $\nu \circ h = g\circ h$. We must show that there is a unique continuous map $h'\colon Z\to X$ with $h=f\circ h'$. Since $f$ is injective (by Stefan's argument) the "uniqueness" part is clear; the question is whether such a map $h'$ exists.
Let $z\in Z$. We have $h(z)\in f(X)$, because if this would not be the case, then $h(z)\in Y\backslash f(X)$, and so $(g\circ h)(z)(h(z))=g_{h(z)}(h(z))=1\neq 0=(\nu\circ h)(z)(h(z))$, which contradicts $\nu\circ h = g\circ h$.
Thus (since $f$ is injective) there is a unique $x\in X$ with $f(x)=h(z)$. Hence we may define a map $h'\colon Z \to X$ by $f(h'(z))=h(z)$ for all $z\in Z$.
It remains to be shown that $h'\colon Z \to X$ is continuous. Let $C\subseteq X$ be closed subset. It suffices to show that ${h'}^{-1}(C)$ is closed.  Since $f$ is injective, we have $f^{-1}(f(C))=C$. Since $h$ is continuous, and $f(C)$ is closed (as we saw in the second paragraph), we see that
$${h'}^{-1}(C) \ =\  {h'}^{-1}(f^{-1}(f(C)))
\ = \ (f\circ h')^{-1}(f(C))\ = \ h^{-1}(f(C))
$$
is closed. Thus $h'$ is continuous. Hence $f$ is an equalizer.
