Suppose we have a finite $\sigma$ -field S, and that sets A and B $\in$ S. Since S is closed under union & complementation [by def] $(A' \cup B')' = (A \cap B)'$ and $(A \cap B)$ lie in S as well. This means that the intersection of any 2 sets in S lie in S as well.

Doesn't this make S a topology as well? So does it follow that every finite $\sigma$ -field is a topology?

Suppose we have a finite $\sigma$ -field $S$, of which $A$ and $B$ are member sets. Since $S$ is closed under union and complementation [by definition], it follows that $(A' \cup B')' = (A \cap B)' \in S$. From closure under complementation, we have that $A \cap B \in S$, implying that $S$ is closed under intersections.

Does it follow that every finite $\sigma$ -field is a topology?