Let $X$ be a separable metric space and $Y$ be a second-countable $\sigma$-compact Hausdorff space. Then the compact-convergence (compact-open) topology on $C(Y,X)$ is metrizable with metric
$$
d(f,g):= \sum\_{n =0}^{\infty} \frac1{2^n} \sup\_{y \in K\_n} \min\left\{d\_X(f(y),g(y)),1\right\},
$$
where $\{K\_n\}\_{n \in \mathbb{N}}$ is a countable compact cover of $Y$ and $d\_X$ is the metric on $X$. Moreover, if $X$ is Banach then $C(Y,X)$ is a Fréchet space.


This is easy to show, but I'm looking for a reference to this result. Thanks in advance. 



Metrizability of topology of compact convergence
According to Engelking (exercise 3.4E, which is based on a [paper](https://www.jstor.org/stable/1969087?seq=1) by Arens):



> 
> If $C(X,\Bbb R)$ (with the compact-open topology and $X$ Tychonoff) is first countable, then $X$ is hemicompact.
> 
> 
> 


A Hausdorff space $X$ is hemicompact if there is a countable family $K\_n$ of compact subsets of $X$ such that every compact $K \subseteq X$ is a subset of some $K\_n$ (i.e. all compacta of $X$ ordered under inclusion has countable cofinality). For $X$ second-countable, hemicompactness is equivalent to local compactness. 


So a space like $\Bbb Q$, which is not locally compact but is $\sigma$-compact has $C(X,\Bbb R)$ not even first countable, let alone metrisable.


But Arens showed in that same paper (ex. 4.2H in Engelking) that for hemicompact $X$ and metrisable $Y$ , $C(X,Y)$ in the compact-open topology *is* metrisable, using a metric like yours. 


So the moral is: you need to add "locally compact" to your $Y$ (and the space then becomes hemicompact and all is well).


