Let $X$ be a topological space and $Y$ be a metric space, $A \subseteq X$ compact. We denote $\mathcal{T}_u^A$ for the topology of the uniform convergence on $C(A,Y)$.
Consider the source:
$$(r_A: C(X,Y) \to (C(A,Y), \mathcal{T}_u^A): f \mapsto f\vert_A)_{A \subseteq X \mathrm{ \ compact}}$$
And define $\mathcal{T}_{uc}$ as the initial topology for this source.
My book then claims the following:
Let $X$ be a Hausdorff space and $Y$ be a metric space. Then $\mathcal{T}_c = \mathcal{T}_{uc}$ where $\mathcal{T}_c$ is the compact-open topology on $C(X,Y)$
It provides the following proof:
Proof: This is an immediate consequence of the definition of initial structures and the fact that if $X$ is compact Hausdorff and $Y$ is metrizable, then $\mathcal{T}_u = \mathcal{T}_c \square$
I can't see why this is true. I can see that from that fact it follows that for $A \subseteq X$ compact it follows that $\mathcal{T}_c^A = \mathcal{T}_u^A$ (since subspaces of Hausdorff spaces are Hausdorff), but I can't see how the result follows from "the definition of initial structures".
First the required background on initial topologies: If $\{f_i: X \to (Y_i, \mathcal{T}_i): i \in I\}$ is a source, then a subbase for the initial topology is given by $\{f_i^{-1}[O]: O \in \mathcal{T}_i, i \in I\}$.
In fact, if $\mathcal{S}_i$ is a given (sub)base for $(Y_i, \mathcal{T}_i)$, then we can also use $$\{f_i^{-1}[O] : O \in \mathcal{S}_i, i \in I\}$$
as a subbase for the initial topology, as can easily be checked, using the definition of initial topology and properties of inverse images.
Now to your source $r_A: C(X,Y) \to (C(A, Y), \mathcal{T}^A_u)$: we can use that $\mathcal{T}^A_u = \mathcal{T}^A_c$ as $A$ is compact Hausdorff and $Y$ is metric, and so we can use the standard compact-open subbase $\{[K,O]^A: K \subseteq A \text{ compact }, O \subseteq Y \text{ open }\}$ for $\mathcal{T}^A_c$ as a subbase for the uniform topology $\mathcal{T}^A_u$, where $[K,O]^A = \{f \in C(A,Y): f[K] \subseteq O\}$. (the superscript $A$ is unusual and only for didactic purposes).
So using these subbases and the aforementioned fact for initial topologies, a subbase for $\mathcal{T}_{uc}$ then becomes
$$\{(r_A)^{-1}[[K,O]^A]: A \subseteq X \text{ compact}, K \subseteq A \text{ compact }, O \subseteq Y \text{ open} \}$$
Now note that by the definitions $r_A^{-1}[[K,O]^A] = [K,O] := \{f \in C(X,Y): f[K] \subseteq O\}$ (clearly, $f|_A[K] = f[K]$ when $K \subseteq A \subseteq X$) where the latter is a standard subbasic element of $\mathcal{T}_c$, and that the above subbase then just becomes (cutting out the "middle men" $A$, as every $K$ is just a compact subset of $X$ too, and we can use $A=K$ for the other direction):
$$\{[K,O]: K \subseteq X \text{ compact}, O \subseteq Y \text{ open}\}$$
which is just the standard subbase for $\mathcal{T}_c$. Topologies with the same subbase are the same, by definition.