= Solution
For every $a\in\mathbb R$, the nonempty order interval
$$
U_a=\langle(a,-1),(a,1)\rangle=\{a\}\times(-1,1)
$$
is open in the <lexicographic order topology on the real plane>. The sets $U_a$ are pairwise disjoint.
Suppose $\{B_n:n\in\mathbb N\}$ were a countable <basis of a topology>. For each $a$, the <basis of a topology> property gives a member containing $(a,0)$ and contained in $U_a$. Let $n(a)$ be the least index of such a member. The same nonempty $B_n$ cannot lie in two disjoint $U_a$, so $a\mapsto n(a)$ is an injection from the uncountable set $\mathbb R$ into $\mathbb N$, a contradiction. \b[This topology is not second countable.]
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