The topology generated by a basis is the intersection of all topologies containing that basis.

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 $U=\bigcup_{i\in I}A_i\in \bigcap_{a\in A}\tau_a$.

i may be missing something, but isn't it true that $$ \tau_{\mathcal{A}} \in \{\tau_{\alpha}\} \Rightarrow \cap_{a\in\alpha}\tau_a \subset \tau_{\mathcal{A}} $$