 Lexicographic order topology on the real plane

Give ${\mathbb{R}}^2$ the lexicographic order and the order topology. Every point $(a,b)$ has an interval neighbourhood between $(a,b-\epsilon )$ and $(a,b+\epsilon )$, lying entirely in the vertical line $\{a\}\times \mathbb{R}$. These intervals give each such line its usual real topology, and every line is open. Thus this space is a disjoint union of uncountably many open copies of the real line and is not a second-countable space: any basis of a topology must supply a different member inside each of these disjoint nonempty open sets.