Arzela-Ascoli compact-open topology If we consider the following theorem

we note that $\mathcal{F}$ is compact and $f_n\in\mathcal{F}$, my question is why the sequence $f_n$ admits a convergent subsequence.
note: If  $\mathcal{F}$ is compact, that does not imply that $\mathcal{F}$ is sequentially compact.
Any hint would be appreciated.
 A: This statement is in fact incorrect.  For instance, suppose $X$ is $\{0,1\}^\mathbb{N}$ with the discrete topology and $Y=\{0,1\}$.  Then $C(X,Y)$ consists of just all maps $X\to Y$, and the compact-open topology is the product topology on $Y^X$ which is compact since $Y$ is compact.  So you can just take $\mathcal{F}$ to be all of $C(X,Y)$.  Define a sequence $(f_n)$ in $C(X,Y)$ by $f_n(s)=s(n)$.  Then this has no convergent subsequence, since for any infinite $A\subseteq \mathbb{N}$ there exists $s\in X$ such that $s|_A$ is not eventually constant, and so the subsequence of $(f_n)$ given by $A$ will not converge at $s$.
However, if you additionally assume $X$ is $\sigma$-compact, then the claimed statement is true because the compact-open topology on $C(X,Y)$ is metrizable.  Indeed, let $d_Y$ be any bounded metric on $Y$ and let $X=\bigcup_n K_n$ where each $K_n$ is compact and the interiors of the $K_n$ cover $X$ (you can choose such $K_n$ since $X$ is both locally compact and $\sigma$-compact).  Then the compact-open topology on $C(X,Y)$ is induced by the metric $$d(f,g)=\sum_n\frac{\sup\{d_Y(f(x),g(x)):x\in K_n\}}{2^n}.$$  (The proof is a bit complicated, but the key fact to use is that any compact subset of $X$ is covered by finitely many of the $K_n$.)
