Generated Topology is a Topology

Theorem
Let $$X$$ be a set, $$\mathcal{S} \subseteq \mathcal{P}(X)$$ and $$\mathcal{T}_\mathcal{S}$$ be the generated topology for $$\mathcal{S}$$.

Then $$\mathcal{T}_\mathcal{S}$$ is a topology on $$X$$.

Proof
We show that $$\mathcal{S}^* = \left\{{\bigcap S | S \subseteq \mathcal{S} \text{ finite}}\right\}$$ (cf. the definition of the generated topology) is a basis.

To see that $$\mathcal{S}^*$$ is a basis, we need to prove two things:
 * 1) $$X = \bigcup \mathcal{S}^*$$
 * 2) For any $$U_1, U_2 \in \mathcal{S}^*$$ and $$x \in U_1 \cap U_2$$ there is a $$U \in \mathcal{S}^*$$ such that $$x \in U \subseteq U_1 \cap U_2$$

First note that $$X = \bigcap \emptyset \in \mathcal{S}^*$$ and therefore $$\bigcup \mathcal{S}^* = X$$.

Additionally, if $$U_1, U_2 \in \mathcal{S}^*$$, then there exist finite sets $$S_1, S_2 \subseteq \mathcal{S}$$ such that $$U_1 = \bigcap S_1$$ and $$U_2 = \bigcap S_2$$ by the definition of $$\mathcal{S}^*$$.

Thus we have:
 * $$U_1 \cap U_2 = \bigcap (S_1 \cup S_2)$$

Because $$S_1 \cup S_2$$ is again a finite set it follows that $$U_1 \cap U_2 \in \mathcal{S}^*$$.

This implies (2).