I'm trying to prove (directly) the existence of equalizers in the category of topological spaces.
First let $s,t:X\to Y$ be continuous maps of topological spaces. Let $E=\{x\in X:s(x)=t(x)\}$ and $i:E\to X$ the inclusion map. $E$ has the subset topology, and $i$ is continuous.
Suppose $f: A\to X$ is a continuous map of topological spaces s.t. $sf=tf$. There is a map of sets $\bar f: A\to E, \bar f(a)=f(a)$ such that $f=i\bar f$. The problem is to show that $\bar f$ is continuous.
Let $U\subset E$ be open (so $U=E\cap V$ where $V\subset X$ is open, and also $U=i^{-1}(V)$). We need to prove that $\bar f^{-1}(U)\subset A$ is open. We have $$\bar f^{-1}(U)=\{a\in A:\bar f(a)\in U\}\\=\{a\in A:f(a)\in U\}\\=f^{-1}(U)$$
If we knew that $U\subset X$ is open, then $f^{-1}(U)\subset A$ would be open, and hence $\bar f^{-1}(U)\subset A$ would be open. But why is $U$ open in $X$ (and is it)? We only know it's open in $E$.
So we have $f: A \to X$ such that $sf=tf$. This means that actually as sets $f[A] \subseteq E$. So define $g: X \to E$ by $g(a)=f(a)$ for all $a \in A$ and this is a well-defined function and $ig=f$ is trivial. It is continuous as a co-domain restriction of a continuous map: $O$ is open in $E$ iff it is of the form $i^{-1}[U]$ for $U$ open in $X$ and $g^{-1}[O]= g^{-1}[i^{-1}[U]] = (ig)^{-1}[U]=f^{-1}[U]$ so $g$ is continuous. Note that I just use the same $g$ as the equalizer in $\mathsf{Set}$ and use only that $i$ induces the initial topology on it to show its continuity ( a map $g$ into $E$ is continuous iff $i \circ g$ is continuous).