Let $e:E\to A$ be a categorical equalizer of $f,g:A\to B$, and let $D$ be the set-theoretical equalizer. What we want to show is that the image of $e$ is $D$.
To see this, let $d:D\hookrightarrow A$ be the obvious inclusion. Since $f\circ d=g\circ d$, by the universal property of $e$, $d$ must factor through it - there is some $c:D\to E$ such that $d=e\circ c$. Since $d$ is onto $D$, so must be $e$.