= Solution
The <lexicographic order> is a <total order>. Each point $p=(a,b)$ lies in the order interval between $(a,b-1)$ and $(a,b+1)$, so the order intervals cover $\mathbb R^2$.
If $p$ lies in both $\langle x_1,y_1\rangle$ and $\langle x_2,y_2\rangle$, let $x=\max(x_1,x_2)$ and $y=\min(y_1,y_2)$ in that <total order>. Then $x<p<y$ and
$$
\langle x_1,y_1\rangle\cap\langle x_2,y_2\rangle=\langle x,y\rangle.
$$
In particular an order interval contains $p$ and is contained in the intersection. Both conditions of part (a) hold, proving that these intervals form a <basis of a topology>. \b[The resulting topology is the <lexicographic order topology on the real plane>.]
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