At the moment I know the answer to the Question (but not to the Problem). First a definition.

A subset $D$ of a topological space $X$ is called *$k$-dense* in $X$ if each compact subset $K\subset X$ can be enlarged to a compact set $\tilde K$ such that $D\cap\tilde K$ is dense in $\tilde K$.

A topological space $X$ is called *$k$-separable* if it contains a countable $k$-dense subset.

It is easy to see that each dense subset of a metrizable space is $k$-dense. So, metrizable separable spaces are $k$-separable.

Also each dense subset of the Sorgenfrey line $\mathbb S$ is $k$-dense, so $\mathbb S$ is $k$-separable, too.

**Theorem.** For any $k$-separable space $X$ and cometrizable space $Y$, the function space $C_k(X,Y)$ with compact-open topology is cometrizable.

*Proof.* Let $\tau$ be the weaker metrizable topology on $Y$ witnessing that $Y$ is cometrizable. Denote by $Y_\tau$ the metrizable topological space $(Y,\tau)$. Let $D$ be a countable $k$-dense subset of $X$. Consider the restriction operator $R:C_k(X,Y)\to Y_\tau^D$, $R:f\mapsto f{\restriction}D$. Let $\sigma$ be the metrizable topology on $C_k(X,Y)$ such that the map $R:(C_k(X,Y),\sigma)\to Y^D_\tau$ is a topological embedding. It can be shown that the topology $\sigma$ witnesses that the function space $C_k(X,Y)$ is cometrizable. $\square$

Since the Sorgenfrey line $\mathbb S$ is $k$-separable, Theorem implies

**Corollary.** The function space $C_k(\mathbb S,\mathbb R)$ is cometrizable.

Now it remains to observe that the Sorgenfrey line $\mathbb S$ admits a topological embedding to its own function space $C_k(\mathbb S,\mathbb R)$ (which is a cometrizable Abelian topological group).