FALSE: the compact-open topology on $C(X,Y)$ is metrizable for every metric $X$ and $Y$

$\rho$ is a metric on $\mathbb{R}$: it is symmetric and vanishes exactly on the diagonal by definition, and for the triangle inequality $\rho(u,w) \le \rho(u,v) + \rho(v,w)$ either $u = w$, when the left side is $0$, or $u \ne w$, when $v$ differs from at least one of $u$ and $w$ and the right side is at least $1 = \rho(u,w)$.

In $X = (\mathbb{R},\rho)$ every subset is open, since $B_\rho(u,1) = \{u\}$ for every $u$; consequently every function $\mathbb{R} \to \mathbb{R}$ is continuous as a map $X \to Y$, so $C(X,Y) = Y^{X} = \mathbb{R}^{\mathbb{R}}$ as sets.

A subset $K \subseteq X$ is compact exactly when it is finite: the family $\{\, \{u\} : u \in K \,\}$ is a family of open subsets covering $K$, so compactness forces finitely many singletons to cover $K$, and conversely every finite set is compact.

For finite $K = \{x_0,\dots,x_{n-1}\}$ and open $V \subseteq \mathbb{R}$ one has $S(K,V) = \pi_{x_0}^{-1}[V] \cap \dots \cap \pi_{x_{n-1}}^{-1}[V]$, and $S(\varnothing,V)$ is the whole space; conversely $\pi_x^{-1}[V] = S(\{x\},V)$ with $\{x\}$ compact.

By step 4.1 every subbasic set of the compact-open topology is open in the topology of pointwise convergence and every subbasic set of the topology of pointwise convergence is open in the compact-open topology; so the two topologies on $\mathbb{R}^{\mathbb{R}}$ are equal, and it suffices to show that the topology of pointwise convergence on $\mathbb{R}^{\mathbb{R}}$ is not metrizable.

Let $\mathcal{B}$ be any at most countable neighbourhood base at $\mathbf{0}$ in that topology; $\mathcal{B}$ is nonempty, since the whole space is a neighbourhood of $\mathbf{0}$ and must contain a member of $\mathcal{B}$, so there is a function $k \mapsto N_k$ with domain $\mathbb{N}$ whose set of values is $\mathcal{B}$.

For $k \in \mathbb{N}$ put $G_k := \{\, x \in \mathbb{R} : \pi_x[N_k] \ne \mathbb{R} \,\}$, a set determined by $N_k$ with nothing selected.

Each $G_k$ is finite: $N_k$ is a neighbourhood of $\mathbf{0}$, so it contains a basic set $B = \bigcap_{j<n} \pi_{x_j}^{-1}[V_j]$ with $\mathbf{0} \in B$, whence $0 \in V_j$ for every $j < n$; for $x$ outside the finite set $F := \{x_0,\dots,x_{n-1}\}$ and any $t \in \mathbb{R}$ the function agreeing with $\mathbf{0}$ everywhere except at $x$, where it takes the value $t$, lies in $B \subseteq N_k$ and has $t$ as its coordinate at $x$, so $\pi_x[N_k] = \mathbb{R}$ and $x \notin G_k$; hence $G_k \subseteq F$ and $G_k$ is finite, a subset of a finite set being finite.

Therefore $G := \bigcup_{k \in \mathbb{N}} G_k$ is at most countable, being a union over $\mathbb{N}$ of at most countable sets; this step and only this step uses the Axiom of Countable Choice.

$\mathbb{R} \setminus G \ne \varnothing$: otherwise $\mathbb{R} \subseteq G$ would make $\mathbb{R}$ at most countable, contradicting its uncountability; so fix $x_{\ast} \in \mathbb{R} \setminus G$.

The set $N := \pi_{x_{\ast}}^{-1}[(-1,1)]$ is a subbasic open set containing $\mathbf{0}$, hence a neighbourhood of $\mathbf{0}$; and no $N_k$ is contained in $N$, since $N_k \subseteq N$ would give $\pi_{x_{\ast}}[N_k] \subseteq (-1,1) \ne \mathbb{R}$ and hence $x_{\ast} \in G_k \subseteq G$, which step 10.1 excludes.

So $\mathcal{B}$ is not a neighbourhood base at $\mathbf{0}$ after all; as $\mathcal{B}$ was an arbitrary at most countable family of neighbourhoods of $\mathbf{0}$, the space has no at most countable neighbourhood base at $\mathbf{0}$ and is not first countable, hence not metrizable.

With step 5.1 this exhibits metric spaces $X$ and $Y$ for which the compact-open topology on $C(X,Y)$ is not metrizable, so the claim is false.