FALSE: the evaluation map on $C(X,Y)$ with the compact-open topology is continuous for every metric $X$

Suppose the claim holds; in particular, with $X:=\mathbb{Q}$ and $Y:=\mathbb{R}$, the evaluation map $e:C(\mathbb{Q},\mathbb{R})\times \mathbb{Q}\to \mathbb{R}$ is continuous.

The constant function $z$ is continuous, and $e(z,0)=0\in V$ with $V$ open in $\mathbb{R}$.

Continuity of $e$ at the point $(z,0)$ gives, by the two bases of [L7], a natural $n$, compact sets $K_0,\dots ,K_{n-1}\subseteq \mathbb{Q}$, open sets $V_0,\dots ,V_{n-1}\subseteq \mathbb{R}$ and an open $U\subseteq \mathbb{Q}$ with $0\in U$, such that $z\in O:=S(K_0,V_0)\cap \cdots \cap S(K_{n-1},V_{n-1})$ and $e[O\times U]\subseteq V$.

Fix a real $\epsilon >0$ with $B_{\mathbb{Q}}(0,\epsilon )\subseteq U$.

Put $K:=K_0\cup \cdots \cup K_{n-1}$, with $K:=\mathrm{\varnothing }$ when $n=0$; each $K_j$ is a compact subset of $\mathbb{R}$ and hence closed in $\mathbb{R}$, so $K$ is a subset of $\mathbb{Q}$ closed in $\mathbb{R}$.

$B_{\mathbb{Q}}(0,\epsilon )\subseteq ̸K$: otherwise $(-\epsilon ,\epsilon )\cap \mathbb{Q}\subseteq K$, and this set is nonempty since it contains $0$; every $t\in (-\epsilon ,\epsilon )$ then satisfies $d(t,(-\epsilon ,\epsilon )\cap \mathbb{Q})=0$, because for a real $s>0$ the set $(\max \{t-s,-\epsilon \},\min \{t+s,\epsilon \})$ is a nondegenerate open interval and so contains a rational within $s$ of $t$; hence $t$ lies in the closure of that set, which the closed $K$ contains, so $(-\epsilon ,\epsilon )\subseteq K\subseteq \mathbb{Q}$, making the uncountable interval $(-\epsilon ,\epsilon )$ at most countable, which is false.

Fix $q\in B_{\mathbb{Q}}(0,\epsilon )$ with $q\mathrm{\notin }K$.

Put $\eta :=\min \{d(q,K),1\}$ when $K\mathrm{\ne }\mathrm{\varnothing }$ and $\eta :=1$ when $K=\mathrm{\varnothing }$; then $0<\eta \le 1$, because $K$ closed in $\mathbb{R}$ and $q\mathrm{\notin }K$ give $q\mathrm{\notin }\bar{K}$ and hence $d(q,K)\mathrm{\ne }0$, while $d(q,K)\ge 0$; and $\mathrm{\mid }q-t\mathrm{\mid }\ge d(q,K)\ge \eta$ for every $t\in K$.

$q\in U$, since $\mathrm{\mid }q-0\mathrm{\mid }<\epsilon$ puts $q$ in $B_{\mathbb{Q}}(0,\epsilon )\subseteq U$.

Put $A:=\{s\in \mathbb{Q}:\mathrm{\mid }s-q\mathrm{\mid }\ge \eta \}$, which is nonempty because $q+1\in \mathbb{Q}$ and $\mathrm{\mid }(q+1)-q\mathrm{\mid }=1\ge \eta$, and define $g:\mathbb{Q}\to \mathbb{R}$ by $g(s):=(2\mathrm{/}\eta )d(s,A)$.

$g$ is Lipschitz with constant $2\mathrm{/}\eta$, hence continuous, so $g\in C(\mathbb{Q},\mathbb{R})$.

$g$ vanishes on $K$: every $t\in K$ lies in $\mathbb{Q}$ and satisfies $\mathrm{\mid }t-q\mathrm{\mid }\ge \eta$ by step 6.1, so $t\in A$ and $d(t,A)=0$.

$g(q)\ge 2$: every $s\in A$ satisfies $\mathrm{\mid }q-s\mathrm{\mid }\ge \eta$, so $\eta$ is a lower bound of the distances from $q$ to the members of $A$ and $d(q,A)\ge \eta$, whence $g(q)=(2\mathrm{/}\eta )d(q,A)\ge 2$.

$g\in O$: for each $j<n$ with $K_j\mathrm{\ne }\mathrm{\varnothing }$ we have $z\in S(K_j,V_j)$, so $0\in V_j$, and $g[K_j]\subseteq \{0\}\subseteq V_j$ by step 8.2 since $K_j\subseteq K$; for $K_j=\mathrm{\varnothing }$ the condition is vacuous; and for $n=0$ the set $O$ is the whole of $C(\mathbb{Q},\mathbb{R})$.

Hence $(g,q)\in O\times U$ while $e(g,q)=g(q)\ge 2$, so $e(g,q)\mathrm{\notin }(-1,1)=V$, contradicting $e[O\times U]\subseteq V$ of step 2.1; the assumption of step 1.1 is therefore false, and the claim fails for $X=\mathbb{Q}$ and $Y=\mathbb{R}$.