We claim that $\webleft (\text{Eq}\webleft (f,g\webright ),x_{0}\webright )$ is the categorical equaliser of $f$ and $g$ in $\mathsf{Sets}_{*}$. First we need to check that the relevant equaliser diagram commutes, i.e. that we have
\[ f\circ \text{eq}\webleft (f,g\webright )=g\circ \text{eq}\webleft (f,g\webright ), \]
which indeed holds by the definition of the set $\text{Eq}\webleft (f,g\webright )$. Next, we prove that $\text{Eq}\webleft (f,g\webright )$ satisfies the universal property of the equaliser. Suppose we have a diagram of the form
in $\mathsf{Sets}_{*}$. Then there exists a unique morphism of pointed sets
\[ \phi \colon \webleft (E,*\webright )\to \webleft (\text{Eq}\webleft (f,g\webright ),x_{0}\webright ) \]
making the diagram
commute, being uniquely determined by the condition
\[ \text{eq}\webleft (f,g\webright )\circ \phi =e \]
via
\[ \phi \webleft (x\webright )=e\webleft (x\webright ) \]
for each $x\in E$, where we note that $e\webleft (x\webright )\in A$ indeed lies in $\text{Eq}\webleft (f,g\webright )$ by the condition
\[ f\circ e=g\circ e, \]
which gives
\[ f\webleft (e\webleft (x\webright )\webright )=g\webleft (e\webleft (x\webright )\webright ) \]
for each $x\in E$, so that $e\webleft (x\webright )\in \text{Eq}\webleft (f,g\webright )$. Lastly, we note that $\phi $ is indeed a morphism of pointed sets, as we have
\begin{align*} \phi \webleft (*\webright ) & = e\webleft (*\webright )\\ & = x_{0},\end{align*}
where we have used that $e$ is a morphism of pointed sets.