Under choice, the lower-limit line is regular and separable but not second countable and therefore not metrizable

By [L1] the space is regular, and by [L2] the countable set $\mathbb Q$ is dense, so it is separable.

Suppose $(B_n)_{n\in\mathbb N}$ is a basis. For each real $x$, the basis condition for $[x,x+1)$ yields a least-index $B_{m(x)}$ with $x\in B_{m(x)}\subseteq[x,x+1)$.

If $m(x)=m(y)$ and $x<y$, then $x\in B_{m(y)}\subseteq[y,y+1)$, impossible. Thus $x\mapsto m(x)$ injects $\mathbb R$ into $\mathbb N$, contradicting $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874).

Therefore the lower-limit line is not second countable. If it were metrizable, its separability from step 1.1 and [L3] would make it second countable, another contradiction.