Problem 5, sec. 13, in Munkres' Topology, 2nd ed.: How to prove the assertion if $\mathscr{A}$ is a subbasis? Show that if $\mathscr{A}$ is a basis for a topology on $X$, then the topology generated by $\mathscr{A}$ equals the intersection of all topologies on $X$ that contain $\mathscr{A}$.
This is what I've managed. However, the following stumps me:
Prove the same if $\mathscr{A}$ is a subbasis.
That is, how to show that, the topology generated by $\mathscr{A}$ as a subbasis is the same as the intersection of all topologies on $X$ that contain $\mathscr{A}$?
Let $\mathscr{T}$ be the topology that consists of all unions of finite intersections of sets in $\mathscr{A}$, and let $\mathscr{T}^\prime$ be the topology that is the intersection of all topologies that contain $\mathscr{A}$.
Now $\mathscr{A}$ is a collection of subsets of $X$ whose union is all of $X$, and each set $A \in \mathscr{A}$ is in both $\mathscr{T}$ and $\mathscr{T}^\prime$.
Am I right so far?
And if so, then what next?

First, we show that $\mathscr{T}'\subseteq \mathscr{T}$.
Suppose $U \in \mathscr{T}'$. Then $U$ is an element of every topology containing $\mathscr{A}$, including $\mathscr{T}$, which is certainly one such topology. So $U \in \mathscr{T}$, which shows $\mathscr{T}'\subseteq \mathscr{T}$.
To see that $\mathscr{T} \subseteq \mathscr{T}'$:
Let $W \in \mathscr{T}$. Then $W = \bigcup W_i$ with each $W_i$ a finite intersection of sets in $\mathscr{A}$, say $W_i = W_{i1} \cap \dots \cap W_{in}$, with $W_{ij} \in \mathscr{A}$. Since $W_{ij} \in \mathscr{A}$, and $\mathscr{A} \subseteq \mathscr{T}'$, and because $\mathscr{T}'$ is a topology (and thus closed under finite intersections), $W_i \in \mathscr{T}'$, and thus (since $\mathscr{T}'$ is also closed under arbitrary unions) $W \in \mathscr{T}'$, so $\mathscr{T} \subseteq \mathscr{T}'$.
Hence, we see that $\mathscr{T} = \mathscr{T}'$.
