Let $X$ and $Y$ be polish spaces, is it true that the collection of all continuous functions $C(X,Y)$, of $X$ to $Y$, when equipped with the compact-open topology, is second-countable?
My goal is to prove that the collection of all isometries $I(X,Y)$ of $X$ to $Y$, equipped with the pointwise convergence, is separable. I think that both topologies coincides in $I(X,Y)$, and, if the first assertion is true, since $I(X,Y)$ is metrizable, it would be separable.
It's not a answer to your first question, but solves the underlying problem. The exercise 3.4.H of Engelking's General Topology says something interesting about the $C_p(X,Y)$ (the space $C(X,Y)$ equipped with the pointwise convergence topology) and $C_{co}(X,Y)$ (the space $C(X,Y)$ equipped with the compact-open topology): $$\operatorname{nw}(C_p(X,Y))\leq \operatorname{w}(X)\operatorname{w}(Y)\;\text{ and }\operatorname{nw}(C_{co}(X,Y))\leq \operatorname{w}(X)\operatorname{w}(Y)$$ where $\operatorname{nw}(X)$ is the network weight of $X$ (see this) Once you prove that $\operatorname{nw}(C_p(X,Y))\leq \operatorname{w}(X)\operatorname{w}(Y)$, it's easy to see that, for every subspace $S$ of $C_p(X,Y)$, $\operatorname{d}(S)\leq \operatorname{nw}(S)\leq\operatorname{nw}(X)$. So, if $X$ and $Y$ are both second-countable spaces, then $C_p(X,Y)$ and $C_{co}(X,Y)$ are both hereditarily separable.
Let $X$ and $Y$ be polish spaces, is it true that the collection of all continuous functions $C(X,Y)$, of $X$ to $Y$, when equipped with the compact-open topology, is second-countable?
Not necessarily.
Moreover, if $X$ is not $\sigma$-compact zero-dimensional space and $|Y|>1$ then $C(X,Y)$ is even not first countable. For instance, if $X=\Bbb R\setminus\Bbb Q$ is a subspace of $\Bbb R$ endowed with the standard topology; it is not $\sigma$-compact by Baire Theorem, beacuse each compact of the space $X$ is nowhere dense.
Indeed, let $f_0: X\to Y$ be an arbitrary constant map such that $f_0(x)=y_0$ for all $x\in X$. Assume that the space $C(X,Y)$ has a countable base at the point $f_0$.
It is easy to show that it has a countable subbase $$\{M(K_n, U_n): K_n\text{ is a compact subset of $X$ and $U_n$ is open in }Y\} \,,$$ where $M(K,U)= \{f\in C(X,Y): f(K)\subset U\}$.
Since the space $X$ is not $\sigma$-compact, there exists a point $x_0\in X\setminus\bigcup K_n$.
Pick from $Y$ an arbitrary point $y_1$ distinct from $y_0$. We claim that there exist no finite family $\mathcal F$ of the form $\{ M(K_n, U_n):n\in F\}$ such that $f_0\in\bigcap\mathcal F\subset M(\{x_0\}, Y\setminus\{y_1\})$.
Indeed, assume the converse. Put $K=\bigcup\{K_n: n\in F\}$.
Since the set $K$ is compact, it is closed in $X$. Since the space $X$ is zero-dimensional, there exists a disjoint from $K$ closed and open neighborhood $V$ of the point $x_0$.
Define a (continuous) map $f:X\to Y$ by putting $f(x)=y_0$, if $x\in X\setminus V$ and $f(x)=y_1$, if $x\in V$. Then $f\in \bigcap\mathcal F\setminus M(\{x_0\}, Y\setminus\{y_1\} )$ , a contradiction.
We can similarly prove that a case when $X$ is Tychonoff and not $\sigma$-compact and $Y=[0,1]$ is also a counterexample.
