 Solution

For every $a\in \mathbb{R}$, the nonempty order intervalis 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. This topology is not second countable.