Munkres Order Topology example question (Section 14 Example 4)

As you mentioned, let $\mathcal{B}$ be a basis. $U$ is open if and only if for all $x$ in $U$ there exists a $B \in \mathcal{B}$ such that $x \in B \subset U$.

Now for your example, let $U = \{b_1\}$, $x = b_1$. If $\{b_1\}$ was actually open, then there would exists a basis open set $B$ such that $b_1 \in B \subset \{b_1\}$. However all basis open sets containing $b_1$ contains some $a_i$. So it not possible that $B \subset \{b_i\}$. The definition of being open according to the basis of the topology is not fulfilled so $\{b_1\}$ can not be open.

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By the way, I interpreted the question as : how does the bold statement imply $\{b_1\}$ is not open? I assumed you know how to prove the bold statement.