Generated Topology is a Topology

Theorem
Let $$X$$ be a set.

Let $$\mathcal S \subseteq \mathcal P \left({X}\right)$$, where $$\mathcal P \left({X}\right)$$ is the power set of $$X$$.

Let $$\mathcal T_\mathcal S$$ 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).