Let $\mathcal{T}$ be the compact-open topology on $C(X,Y)$. Let $$\mathcal{S}_0 = \{[K,U]:K\subseteq X\text{ is compact and }U\subseteq Y\text{ is open}\},$$ the standard subbase for $\mathcal{T}$, and let $\mathcal{B}_0$ be the associated base, $$\mathcal{B}_0 = \{\bigcap\mathcal{F}:\mathcal{F}\subseteq \mathcal{S}_0\text{ and }\mathcal{F}\text{ is finite}\}.$$
In Lemma $4.1.6$ we’re given a subbase $\mathcal{S}$ of $Y$, and the claim is that $$\mathcal{S}_1 = \{[K,V]:V \in \mathcal{S},K \subseteq X\text{ compact}\}$$ is a subbase for $\mathcal{T}$. Stripped of its details, the argument given there shows that if $$\mathcal{B}_1 = \{\bigcap\mathcal{F}:\mathcal{F}\subseteq \mathcal{S}_1\text{ and }\mathcal{F}\text{ is finite}\}$$ is the base generated by $\mathcal{S}_1$, then for any $[K,U]\in \mathcal{S}_0$ and any $\gamma\in [K,U]$ there is an $H\in \mathcal{B}_1$ such that $\gamma \in H \subseteq [K,U]$. From this it follows immediately that each member of $\mathcal{S}_0$ is a union of members of $\mathcal{B}_1$, and you’re wondering why this is sufficient to show that $\mathcal{S}_1$ is a subbase for $\mathcal{T}$.
To show that $\mathcal{S}_1$ is a subbase for $\mathcal{T}$, we need only show that $\mathcal{B}_1$ is a base for $\mathcal{T}$, i.e., that every member of $\mathcal{T}$ is a union of members of $\mathcal{B}_1$. This will certainly follow if we can show that every member of the base $\mathcal{B}_0$ is a union of members of $\mathcal{B}_1$. Fix $B \in \mathcal{B}_0$, say $$B = [K_1,U_1]\cap [K_2,U_2]\cap\dots\cap [K_n,U_n],$$ where $[K_1,U_1],[K_2,U_2],\dots,[K_n,U_n] \in \mathcal{S}_0$. Suppose that $\gamma\in B$. Then $\gamma\in [K_i,U_i]$ for $i=1,2,\dots,n$, so the proof of Lemma $4.1.6$ guarantees that there are sets $H_i \in \mathcal{B}_1$ such that $\gamma \in H_i \subseteq [K_i,U_i]$ for $i=1,2,\dots,n$. Let $H = \bigcap\limits_{i=1}^n H_i$; then $H\in \mathcal{B}_1$, and $\gamma \in H \subseteq B$. It follows that $B$ is a union of members of $\mathcal{B}_1$, which is exactly what we needed.
(I suppose that I should mention in passing that since $\mathcal{S}_1$ is clearly a subset of $\mathcal{T}$, it cannot generate a topology finer than $\mathcal{T}$; the only question is whether it generates all of $\mathcal{T}$.)
It is a good start but not enough. You should add the following.
If $\mathcal V$ denotes a base of a topology on $X$ and $f:X\to Y$ is a function then $f$ will be open if $f(V)$ is open for every $V\in\mathcal V$.
This because an open set $U$ can be written as a union of elements of $\mathcal V$ and function $f$ respects unions in the sense that $f(\bigcup_{\lambda\in\Lambda}V_\lambda)=\bigcup_{\lambda\in\Lambda}f(V_{\lambda})$.
So if $V_\lambda\in\mathcal V$ for every $\lambda\in\Lambda$ then $f(U)=f(\bigcup_{\lambda\in\Lambda}V_\lambda)=\bigcup_{\lambda\in\Lambda}f(V_{\lambda})$ is a union of open sets, hence is open.
Best Answer
First, $M(X,Y)$ equals $Y^X=\prod_{x\in X} Y$, the set of all functions from $X$ to $Y$ since ever function on a discrete space is continuous.
Now the subbasis for the product topology on $Y^X$ consists of all $$e_x^{-1}(U)=\{(f(x))_{x\in X}\mid e_x(f)=f(x)\in U\}$$ ranging over the open subsets $U$ of $X$ and the elements $x$ of $X$. The subbasis for the compact-open topology on $M(X,Y)$ consists of the sets $$(K,U)=\{f:X\to Y\mid f(x)\in U\forall x\in K\}$$ for all compact $K$ and open $U$.
Can you show that each $e^{-1}_x(U)$ has the form $(K,U)$ for some compact $K$, and conversely, that each $(K,U)$ is an intersection of $e_x^{-1}(U)$ over all $x$ in a finite set $\{x_1,...,x_n\}$ ?