In A First Course in Topology, ch. 4 §7.7, Robert Conover asks the reader to prove that the following set is a basis for some topology on the set $[-1, 1]$: $$ \mathscr B = \{ [-1, b), (a, 1] : -1 < a, b < 1 \} \cup \{ 0 \}\textrm . $$ (It is also asked to show that the singleton set $\{0\}$ is the only singleton set that is open in the topology generated by $\mathscr B$.)
This seems to be impossible. For consider the sets $B_1 = [-1, \frac{2}{3})$ and $B_2 = (\frac{1}{3}, 1]$, both of which are in $\mathscr B$. If $\mathscr B$ is a basis for a topology, then for every point $x \in B_1 \cap B_2$ ($= (\frac{1}{3}, \frac{2}{3})$) there must exist a set $B_3 \in \mathscr B$ such that $x \in B_3 \subseteq B_1 \cap B_2$. But any set in $\mathscr B$ must contain points which are either smaller than $\frac{1}{3}$ or greater than $\frac{2}{3}$, so they cannot be subsets of $B_1 \cap B_2$, so $\mathscr B$ cannot be the basis for a topology.
What is wrong with my argumentation? Or could the book be mistaken?
I assume the correct question should have $$\mathscr{B} = \{[-1, a), (b,1] : -1 < a < 0 < b < 1\} \cup\{\{0\}\}.$$ You are correct that if we allow these sets to cross over $0$ then we obtain intervals like $(-\frac12, \frac12)$ which cannot be covered by elements of $\mathscr{B}$.