= Solution
Take $U,V\in\mathcal T$ and $x\in U\cap V$. Their representations as unions provide $B_1,B_2\in\mathcal B$ with $x\in B_1\subseteq U$ and $x\in B_2\subseteq V$. The intersection condition supplies $B\in\mathcal B$ with $x\in B\subseteq B_1\cap B_2\subseteq U\cap V$.
It follows that $U\cap V$ is the union of all members of $\mathcal B$ contained in it, and hence belongs to $\mathcal T$. Induction gives closure under every finite intersection. Together with part (i), this proves that $\mathcal T$ is a <topology>. Each $B\in\mathcal B$ belongs to $\mathcal T$ as a union with one member, and every open set is a union of such members by construction. \b[Thus $\mathcal B$ is a basis for $\mathcal T$.] Here the printed subset sign is interpreted as inclusion, allowing equality, as in the usual basis criterion.
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