Let $X$ be a compact metric space, ans $Y$ a space with a countable basis.
Let $Y^X$ (here, the set of continuous functions $X\to Y$) be assigned the compact-open topology, i.e. the topology spanned by sets of the form $F(C,U) = \{f\in Y^X \mid f(C)\subset U\}$ where $C\subset X$ is compact and $U\subset Y$ is open.
The exercise I have claims that $Y^X$ has a countable basis, and I have to admit I'm stuck.
What I started doing :
I wanted to see how the $F(C,U)$'s behaved under union, intersection and so on to get a better grip on what a countable basis may be, so I proved (for all $C,C', U, U'$ in the right set) : $F(C,U)\cup F(C,U') \subset F(C, U\cup U')$, $F(C,U)\cap F(C, U') = F(C, U\cap U')$, $F(C,U)\cap F(C',U) = F(C\cup C', U)$, and these can obviously be generalized to infinite unions and intersections.
My thoughts were then to use the compactness of the $C$'s in the $F(C,U)$'s to note that whenever $f\in F(C, \displaystyle\bigcup_{i\in I} U_i)$, then by continuity of $f$, there is a finite $J\subset I$ such that $f\in F(C, \displaystyle\bigcup_{i\in J} U_i)$. But the $J$ may depend on $f$, and so this isn't helpful...
As you can see, I'm stuck. If you could give me indications (without giving away the whole solution if possible) I would gladly hear them.
Fix countable bases in $X$ and $Y$. Take all $C$ that are compact and finite unions of compact closures of basic elements, and show that $F(C,U)$, where $C$ is as aforementioned and $U$ comes from the base for $Y$, form a (countable) base for $C(X,Y)$.
To flesh it out a bit more: As $X$ has a countable base (being metrisable compact) we can find a countable base $\mathscr{B}$ for $X$ that is closed under finite unions, and a countable base $\mathscr{B}'$ for $Y$ also closed under finite intersections. (A countable set has countably many finite subsets whose union we add to the family).
Now define $\mathscr{B''} = \{F(\overline{B_1}, B_2): B_1 \in \mathscr{B}, B_2 \in \mathscr{B'} \}$, which are open subsets of the compact open topology on $C(X,Y)$ (closed sets in $X$ are compact).
If $f \in F(C,O)$, we need to find a set of the form $F(\overline{B_1}, B_2)$ that contains $f$ and is contained in $F(C,O)$. Try to do this using that $f[C] \subset O$ is compact so that $f[C]$ can be covered by one $\mathscr{B'}\ni B_2 \subset O$ (Using compactness and closedness under finite intersections). Now cover $C$ by elements of $\mathscr{B}$ whose closures sit inside $B_2$. A finite subcover (and using $\mathscr{B}$ is closed under finite unions ) does the rest..