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? 



Finite Topology vs sigma Field
Yes, it is also a topology on its union, the largest member of $S$. Since $S$ is finite, the arbitrary union rquirement amounts to finite union, which you have. 


In fact,$S$ is a Boolean algebra, and since it is finite, it is isomorphic to a powerset algebra---the power set of the atoms of $S$ (the minimal non-empty elements of $S$). Every set in $S$ is a union of a finite number of these atoms. 


If these atoms are singletons, then $S$ will be the discrete topology on the underlying set. If not, however, then $S$ will clearly not separate those points, and so will not be Hausdorff and so on.


