[proofplan]
We define the object part using local smallness, so each hom-class $\operatorname{Hom}_{\mathcal C}(X,A)$ is a set. For a morphism $f: X \to Y$ in $\mathcal C$, we define the corresponding map by precomposition with $f$. We then verify that precomposition by an identity morphism is the identity function, and that precomposition respects composition in the order required by the opposite category.
[/proofplan]
[step:Define the object and morphism assignments]
For each object $X$ of $\mathcal C$, define
\begin{align*}
F(X) := \operatorname{Hom}_{\mathcal C}(X,A).
\end{align*}
Since $\mathcal C$ is locally small, $F(X)$ is a set. Thus $F$ is an object assignment from $\mathcal C^{\operatorname{op}}$ to $\mathbf{Set}$.
Let $f: X \to Y$ be a morphism in $\mathcal C$. This determines a morphism $f^{\operatorname{op}}: Y \to X$ in $\mathcal C^{\operatorname{op}}$. Define
\begin{align*}
F(f^{\operatorname{op}}): F(Y) &\to F(X) \\
g &\mapsto g \circ f.
\end{align*}
This is well-defined because if $g \in F(Y)$, then $g: Y \to A$ in $\mathcal C$, and therefore $g \circ f: X \to A$ is a morphism in $\mathcal C$, so $g \circ f \in F(X)$.
[/step]
[step:Verify preservation of identity morphisms]
Let $X$ be an object of $\mathcal C$. The identity morphism $\operatorname{id}_X: X \to X$ in $\mathcal C$ determines the identity morphism $\operatorname{id}_X^{\operatorname{op}}: X \to X$ in $\mathcal C^{\operatorname{op}}$. For every $g \in F(X)$, so $g: X \to A$, associativity and the identity law in $\mathcal C$ give
\begin{align*}
F(\operatorname{id}_X^{\operatorname{op}})(g)
= g \circ \operatorname{id}_X
= g.
\end{align*}
Hence $F(\operatorname{id}_X^{\operatorname{op}}) = \operatorname{id}_{F(X)}$.
[/step]
[step:Verify preservation of composition in the opposite category]
Let $f: X \to Y$ and $h: Y \to Z$ be composable morphisms in $\mathcal C$. In $\mathcal C^{\operatorname{op}}$, the corresponding morphisms are
\begin{align*}
h^{\operatorname{op}} &: Z \to Y, &
f^{\operatorname{op}} &: Y \to X,
\end{align*}
and their composite is
\begin{align*}
f^{\operatorname{op}} \circ_{\mathcal C^{\operatorname{op}}} h^{\operatorname{op}}
= (h \circ_{\mathcal C} f)^{\operatorname{op}}.
\end{align*}
For every $g \in F(Z)$, so $g: Z \to A$, we compute
\begin{align*}
F\bigl(f^{\operatorname{op}} \circ_{\mathcal C^{\operatorname{op}}} h^{\operatorname{op}}\bigr)(g)
&= F\bigl((h \circ_{\mathcal C} f)^{\operatorname{op}}\bigr)(g) \\
&= g \circ_{\mathcal C} (h \circ_{\mathcal C} f) \\
&= (g \circ_{\mathcal C} h) \circ_{\mathcal C} f \\
&= F(f^{\operatorname{op}})(g \circ_{\mathcal C} h) \\
&= \bigl(F(f^{\operatorname{op}}) \circ_{\mathbf{Set}} F(h^{\operatorname{op}})\bigr)(g).
\end{align*}
The third equality is associativity of composition in $\mathcal C$. Therefore
\begin{align*}
F\bigl(f^{\operatorname{op}} \circ_{\mathcal C^{\operatorname{op}}} h^{\operatorname{op}}\bigr)
= F(f^{\operatorname{op}}) \circ_{\mathbf{Set}} F(h^{\operatorname{op}}).
\end{align*}
Thus $F$ preserves composition.
[/step]
[step:Conclude that the assignment is a functor]
The assignment $F$ sends every object of $\mathcal C^{\operatorname{op}}$ to a set, sends every morphism of $\mathcal C^{\operatorname{op}}$ to a function between the corresponding sets, preserves identity morphisms, and preserves composition. Hence $F$ is a functor
\begin{align*}
F = \operatorname{Hom}_{\mathcal C}(-,A): \mathcal C^{\operatorname{op}} \to \mathbf{Set}.
\end{align*}
[/step]
