 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. 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.