Consider the intersection of all topologies on $x$ that contain $\mathcal{A}$ (we will call this set $T$). Let us also consider the topology generated by $\mathcal{A}$, which we will call $T_{\mathcal{A}}$. We wish to prove that $T \ = \ T_{\mathcal{A}}$. First, we demonstrate that $T_{\mathcal{A}} \ \subset \ T$. Consider some element $U_{\mathcal{A}}$ of $T_{\mathcal{A}}$. As we proved previously, every element of the topology generated by $\mathcal{A}$ will be a union of some collection of elements in $\mathcal{A}$. Since it is true that $\mathcal{A} \ \subset \ T$, and it is also true that an intersection of topologies will also be a topology, it follows that any union of elements of $\mathcal{A}$ will be contained in $T$ (by definition of a topology). This implies that $U_{\mathcal{A}} \ \in \ T$ and $T_{\mathcal{A}} \ \subset \ T$.

A practically identical argument follows for the case of $\mathcal{A}$ being a subbasis.

We will now prove that $T \ \subset \ T_{\mathcal{A}}$. Consider some general element $U$ of $T$. By definition of $T$, this element will be contained in every topology on $X$ that contains $\mathcal{A}$. Well, it is clear that $T_{\mathcal{A}}$ is a topology on $X$ that contains $\mathcal{A}$, so it follows that $U \ \in \ T_{\mathcal{A}}$, and $T \ \subset \ T_{\mathcal{A}}$. Proving this same fact for the case of $\mathcal{A}$ being a subbasis follows from an identical argument.

From these two inclusions, it follows that $T \ = \ T_{\mathcal{A}}$.