Assuming choice, the lower-limit plane is first countable, separable, and ccc, but not second countable or Lindelöf

Let $S$ be the lower-limit line. Under Choice, the countable-product theorem makes $S^2$ first countable, and the rational grid is a countable dense subset; hence $S^2$ is separable and therefore ccc. The antidiagonal $A=\{(x,-x):x\in\mathbb R\}$ is an uncountable closed discrete subspace, as proved in Refuted: separability is hereditary and Assuming countable choice, refuted: Lindelöfness is productive. If $S^2$ were second countable, hereditary second countability would make the discrete space $A$ second countable, which is impossible because every basis of a discrete space contains all its singletons. The explicit open cover in Assuming countable choice, refuted: Lindelöfness is productive shows directly that $S^2$ is not Lindelöf.