The topology generated by a basis is the intersection of all topologies containing that basis. This question is from Munkres' Topology, section 13, exercise 5. I ask for verification and/or comments upon mistakes and inaccuracies.

Let $\mathcal{A}$ be a basis for a topology on $X$. We are to show that the topology generated by $\mathcal{A}$ is the intersection of all topologies on $X$ containing $\mathcal{A}$.

Here is my claimed solution:

Let $\{\tau_{\alpha}\}$ be the family of topologies containing $\mathcal{A}$. Consider the incersection $\bigcap_{a \in \alpha}\tau_{a}$. It is clear that this intersection contains $\mathcal{A}$, and moreover, is a topology, since for any two elements that are shared, intersections and unions must be shared, by the definition of a topology.
We proceed to show that $\bigcap_{a\in\alpha}\tau_a \subset \tau_{\mathcal{A}}$. Let $U$ be an arbitrary element of $\bigcap_{a\in\alpha}\tau_a$. We see that $U$ must either be in $\mathcal{A}$ or be a union or intersection of elements of $\mathcal{A}$, since $\bigcap_{a\in\alpha}\tau_a$ is clearly the largest topology containing every topology containing $\mathcal{A}$, and by the definition of a topology, this must be exactly the elements of $\mathcal{A}$ and unions and intersections therein. $\bigcap_{a\in\alpha}\tau_a \supset \tau_{\mathcal{A}}$ follows directly.

What I am not sure about here is "...by the definition of a topology, this must be exactly the elements of $\mathcal{A}$ and unions and intersections therein." Since the intersection is the largest topology containing every $\tau_\alpha$, any other element could imply new elements by unions and intersections, creating a new topology not necessarily contained in every $\tau_\alpha$. While I am somewhat confident that my intuition here is correct, I feel that my argument lacks a certain precision.
 A: The argument is fine, though it is phrased a little bit unclearly. Here is a way to make it more precise. If $U\in \tau_{\mathcal{A}}$, then it is a union of basis elements, say $U=\bigcup_{i\in I}A_i$, where $A_i \in \mathcal{A}$ for all $i$. Since $\bigcap_{a\in A}\tau_a$ contains $\mathcal{A}$, we have $A_i\in \bigcap_{a\in A}\tau_a$ for all $i$ and since $\bigcap_{a\in A}\tau_a$ is a topology, we also have $U=\bigcup_{i\in I}A_i\in \bigcap_{a\in A}\tau_a$.
A: i may be missing something, but isn't it true that
$$
\tau_{\mathcal{A}} \in \{\tau_{\alpha}\} \Rightarrow \cap_{a\in\alpha}\tau_a \subset \tau_{\mathcal{A}}
$$
A: For those who hate interpret symbolic proofs:
Each open set in the generated topology is union of basis elements and each basis elements are in any containing topology meaning that any of their union is in the said topology. This shows generated topology is contained in any topology that contains its basis and thus contained in their intersection. Equality must holds for the generated topology is also a topology that contains its basis so intersection cannot be larger than it. 
