Metric Space is Lindelöf iff Second-Countable

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Then $M$ is Lindelöf $M$ is second-countable.

Sufficient Condition
We have from Second-Countable Space is Lindelöf that second-countability implies Lindelöf in all topological spaces, regardless of whether they are metric spaces or not.

Necessary Condition
Suppose $M = \left({A, d}\right)$ is Lindelöf.

Let us define the open covers on $A$:
 * $\mathcal C_k = \left\{{N_{1/k} \left({x}\right): x \in S}\right\}$

for all $k \in \N_{>0}$.

As $M$ is Lindelöf, each one of these has a countable subcover.

The union of all these subcovers is a countable basis for the topology on $A$.

Hence the result, by definition of second-countable space.