Assuming countable choice, a metrizable space is second countable if and only if it is separable if and only if it is Lindelöf

Suppose $D$ is at most countable and dense. If $D=\varnothing$, then $X=\varnothing$ and the empty family is a basis. Otherwise the family $\mathcal B=\{B(d,1/n):d\in D,\ n\ge1\}$ is at most countable. It is a basis: if $x\in U$ with $U$ open, choose $\varepsilon>0$ with $B(x,\varepsilon)\subseteq U$, then choose $n$ with $2/n<\varepsilon$ and $d\in D\cap B(x,1/n)$. Now $x\in B(d,1/n)\subseteq B(x,2/n)\subseteq U$. Thus separability implies second countability.

[L1] gives second countable implies Lindelöf.

Suppose $X$ is Lindelöf. For each $n\ge1$, the radius-$1/n$ balls cover $X$. Using [A1], choose an at most countable set $D_n$ of centres whose radius-$1/n$ balls cover $X$. Then $D=\bigcup_{n\ge1}D_n$ is at most countable by [L2]. It is dense: for $x\in U$ open, choose $\varepsilon>0$ with $B(x,\varepsilon)\subseteq U$ and $n$ with $1/n<\varepsilon$; some $d\in D_n$ has $x\in B(d,1/n)$, so $d\in B(x,\varepsilon)\subseteq U$. Thus Lindelöf implies separable.

The three implications prove the equivalence.