 Solution

By the covering condition, $X={\bigcup }_{B\in \mathcal{B}}B$ belongs to $\mathcal{T}$. The empty set belongs to $\mathcal{T}$ as the empty union. An arbitrary union of sets that are themselves unions of members of $\mathcal{B}$ is still a union of members of $\mathcal{B}$. Thus $\mathcal{T}$ contains $X,\varnothing$ and is closed under arbitrary unions, as required for a topology.