[proofplan]
We construct an explicit quasi-inverse to the inclusion functor. For every object of $\mathcal{C}$, choose its unique representative in the skeleton together with an isomorphism to that representative. These choices define a functor $G:\mathcal{C}\to\mathcal{S}$ by conjugating morphisms along the chosen isomorphisms. The fullness of $\mathcal{S}$ ensures that the conjugated morphisms lie in $\mathcal{S}$, and the chosen isomorphisms assemble into a natural isomorphism from $\operatorname{id}_{\mathcal{C}}$ to $I\circ G$.
[/proofplan]
[step:Choose skeletal representatives and comparison isomorphisms]
For each object $C\in \operatorname{Ob}(\mathcal{C})$, since $\mathcal{S}$ is a skeleton of $\mathcal{C}$, there exists a unique object $G(C)\in \operatorname{Ob}(\mathcal{S})$ such that $C$ is isomorphic to $I(G(C))$ in $\mathcal{C}$. Choose an isomorphism
\begin{align*}
\alpha_C:C\longrightarrow I(G(C))
\end{align*}
in $\mathcal{C}$.
For an object $S\in \operatorname{Ob}(\mathcal{S})$, the unique skeletal representative of $I(S)$ is $S$ itself. Thus we choose
\begin{align*}
G(I(S))=S,\qquad \alpha_{I(S)}=\operatorname{id}_{I(S)}.
\end{align*}
[/step]
[step:Define the quasi-inverse functor on morphisms]
Define a functor
\begin{align*}
G:\mathcal{C}\longrightarrow \mathcal{S}
\end{align*}
as follows. On objects, $G$ sends $C$ to the skeletal representative $G(C)$ chosen above. For a morphism $f:C\to D$ in $\mathcal{C}$, define
\begin{align*}
G(f):G(C)\longrightarrow G(D)
\end{align*}
to be the unique morphism of $\mathcal{S}$ whose image under the inclusion $I$ is
\begin{align*}
I(G(f))=\alpha_D\circ f\circ \alpha_C^{-1}:I(G(C))\longrightarrow I(G(D)).
\end{align*}
This morphism exists and is unique because $\mathcal{S}$ is a full subcategory of $\mathcal{C}$, so for all objects $A,B\in \operatorname{Ob}(\mathcal{S})$, the inclusion induces a bijection
\begin{align*}
\operatorname{Hom}_{\mathcal{S}}(A,B)\cong \operatorname{Hom}_{\mathcal{C}}(I(A),I(B)).
\end{align*}
We verify functoriality. For every object $C\in \operatorname{Ob}(\mathcal{C})$,
\begin{align*}
I(G(\operatorname{id}_C))
=\alpha_C\circ \operatorname{id}_C\circ \alpha_C^{-1}
=\operatorname{id}_{I(G(C))},
\end{align*}
so $G(\operatorname{id}_C)=\operatorname{id}_{G(C)}$ by faithfulness of the inclusion on hom-sets.
If $f:C\to D$ and $g:D\to E$ are morphisms in $\mathcal{C}$, then
\begin{align*}
I(G(g\circ f))
=\alpha_E\circ g\circ f\circ \alpha_C^{-1}.
\end{align*}
On the other hand,
\begin{align*}
I(G(g)\circ G(f))
&=I(G(g))\circ I(G(f)) \\
&=(\alpha_E\circ g\circ \alpha_D^{-1})\circ(\alpha_D\circ f\circ \alpha_C^{-1}) \\
&=\alpha_E\circ g\circ f\circ \alpha_C^{-1}.
\end{align*}
Again by faithfulness of the inclusion on hom-sets,
\begin{align*}
G(g\circ f)=G(g)\circ G(f).
\end{align*}
Thus $G:\mathcal{C}\to\mathcal{S}$ is a functor.
[/step]
[step:Show that $G\circ I$ is the identity functor on the skeleton]
Let $S\in \operatorname{Ob}(\mathcal{S})$. By our choice of representatives,
\begin{align*}
(G\circ I)(S)=G(I(S))=S.
\end{align*}
If $h:S\to T$ is a morphism in $\mathcal{S}$, then $\alpha_{I(S)}=\operatorname{id}_{I(S)}$ and $\alpha_{I(T)}=\operatorname{id}_{I(T)}$, so
\begin{align*}
I(G(I(h)))
=\operatorname{id}_{I(T)}\circ I(h)\circ \operatorname{id}_{I(S)}
=I(h).
\end{align*}
Since $I$ is faithful on hom-sets, $G(I(h))=h$. Therefore
\begin{align*}
G\circ I=\operatorname{id}_{\mathcal{S}}.
\end{align*}
[/step]
[step:Assemble the chosen isomorphisms into a natural isomorphism]
The family
\begin{align*}
\alpha_C:C\longrightarrow I(G(C))
\end{align*}
defines the components of a natural transformation
\begin{align*}
\alpha:\operatorname{id}_{\mathcal{C}}\Longrightarrow I\circ G.
\end{align*}
Indeed, for every morphism $f:C\to D$ in $\mathcal{C}$, the definition of $G(f)$ gives
\begin{align*}
I(G(f))\circ \alpha_C
=(\alpha_D\circ f\circ \alpha_C^{-1})\circ \alpha_C
=\alpha_D\circ f.
\end{align*}
This equality is precisely the commutativity condition for the naturality square with top arrow $f:C\to D$, left arrow $\alpha_C:C\to I(G(C))$, right arrow $\alpha_D:D\to I(G(D))$, and bottom arrow $I(G(f)):I(G(C))\to I(G(D))$. Since each $\alpha_C$ is an isomorphism, $\alpha$ is a natural isomorphism
\begin{align*}
\operatorname{id}_{\mathcal{C}}\cong I\circ G.
\end{align*}
[/step]
[step:Conclude that the inclusion is an equivalence]
We have constructed a functor $G:\mathcal{C}\to\mathcal{S}$ such that
\begin{align*}
G\circ I=\operatorname{id}_{\mathcal{S}}
\end{align*}
and
\begin{align*}
\operatorname{id}_{\mathcal{C}}\cong I\circ G.
\end{align*}
Therefore $G$ is a quasi-inverse to the inclusion functor $I:\mathcal{S}\hookrightarrow\mathcal{C}$. Hence $I$ is an equivalence of categories.
[/step]
