Boundedness does not replace pointwise relative compactness for an arbitrary metric target

The function $d$ satisfies the three axioms of [L3]: (M1) holds because $d(m,n)=0$ was defined to mean $m=n$; (M2) holds because the defining cases are symmetric in $m,n$; and for (M3), if $d(m,p)=0$ the inequality is trivial, while if $d(m,p)=1$ then $m\mathrm{\ne }p$, so $n$ differs from at least one of $m,p$ and the right side is at least $1$. So $d$ is a metric. Its metric topology is discrete because $B(n,1)=\{n\}$.

Every map $X\to Y$ is constant. The whole family $\mathcal{F}=C(X,Y)$ is equicontinuous, and $\mathcal{F}(\ast )=Y$ is bounded because it lies in the radius-$2$ ball about $0$.

Evaluation at $\ast$ is a bijection $\mathcal{F}\to Y$. By [L4], the inverse image of each open $V\subseteq Y$ is $S(\{\ast \},V)$, so evaluation is a homeomorphism for the compact-open topology.

The open cover $\{\{n\}:n\in \mathbb{N}\}$ of the infinite discrete space $Y$ has no finite subcover. Thus $Y$ and hence $\mathcal{F}$ are not compact, while the family is equicontinuous and pointwise bounded. Moreover $\overline{\mathcal{F}(\ast )}=Y$ is not compact, displaying exactly the missing hypothesis in [L1].