This is a question from Munkres, $\S34$ Question 8. Hint given is that regularity is essential, and that closed subspace of a Lindelöf space is Lindelöf.
The following is what I have so far: Let $X$ be such space. Since $X$ is locally metrizable, for any point $x\in X$ there exists a neighbourhood $U$ that is metrizable under subspace topology. Since $X$ is also regular, for the aforementioned neighbourhood $U$ of $x$, there exists a closed neighbourhood $\bar{V}$ of $x$ such that $\overline{V}\subseteq U$, so $\overline{V}$ is also metrizable. Since $\overline{V}$ is closed subset of a Lindelöf space $X$, $\overline{V}$ is Lindelöf. Therefore $\overline{V}$ has countable basis.
What I have in my mind is this: since $X$ is Lindelöf, for any open cover of $X$ there exists a countable subcover. If it is possible to let the various $\overline{V}$ be the subcover (not sure how this works, since it is about open cover), then $X$ will have a countable collection of countable basis, therefore it has countable basis. But to be honest I am not sure whether it is even the correct direction. Also I am not sure whether the regularity can be used elsewhere.
Best Answer
Use the sets $V$, not their closures.
You have for each $x\in X$ an open nbhd $V_x$ whose closure has a countable base. Second countability is hereditary, so each $V_x$ also has a countable base $\mathscr{B}_x$, and since $V_x$ is open in $X$, the members of $\mathscr{B}_x$ are open in $X$. $X$ is Lindelöf, so there is a countable $C\subseteq X$ such that $\{V_x:x\in C\}$ covers $X$; let $\mathscr{B}=\bigcup_{x\in C}\mathscr{B}_x$, and show that $\mathscr{B}$ is a base for the topology on $X$. (If you get completely stuck, mouse over the spoiler-protected block below.)