= Solution
Define $\mathcal T$ to be all unions of members of $\mathcal B$, including the empty union. Then $\varnothing\in\mathcal T$ and the covering assumption gives $X\in\mathcal T$. Arbitrary unions of such sets are again unions of basis members.
For two members $U=\bigcup_iB_i$ and $V=\bigcup_jB'_j$, distribute intersection over union:
$$
U\cap V=\bigcup_{i,j}(B_i\cap B'_j).
$$
Each intersection is a union of members of $\mathcal B$ by the second assumption, so $U\cap V\in\mathcal T$. Repeating gives closure under all finite intersections. These are precisely the <topology> axioms. Every $B\in\mathcal B$ belongs to $\mathcal T$, and every set in $\mathcal T$ is a union of such members by construction. Thus
$$
\boxed{\mathcal T=\left\{\bigcup\mathcal C:\mathcal C\subseteq\mathcal B\right\},\qquad
\mathcal B\text{ is a basis for }\mathcal T.}
$$
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