Every Lindelöf topological group is isomorphic to a subgroup of the product of second countable topological groups. I want to show that every Lindelöf topological group is isomorphic to a subgroup of the product of second countable topological groups. I received an answer using the fact that Lindelöf topological groups are $\omega$-narrow, but I want to show it by using the following theorem.
Theorem: Every Hausdorff topological group $G$ is topologically isomorphic to a subgroup of the group of isometries $Is(M)$ of some metric space $M$, where $Is(M)$ is taken with the topology of pointwise convergence.
Any help would be greatly appreciated!

Let $G$ be a Lindelöf group. By your theorem, you can assume that $G$ is a subgroup of $\operatorname{Iso}(M)$ for some metric space $M$. Now, consider the decomposition of $M$ into $G$-orbits, call them $M_a$. Since each orbit is an image of $G$, each $M_a$ is Lindelöf. Furthermore, since each $M_a$ is metrizable, they each have a countable base. 
Now, for every $M_a$, since $G$ acts on $M_a$, there is a natural homomorphism of topological groups from $G$ to $\operatorname{Iso}(M)$. Then, the diagonal product of these homomorphisms gives you the embedding of $G$ into the product of second countable groups.
