Difference between closed under pairwise intersection and closed under finite intersection. What does it mean by pairwise intersection? Is it the same as finite intersection?
First, it is trivial that if a set is closed under finite intersection, it is closed under pairwise as well. Now, I show a set closed under pairwise intersection is also closed under finite intersection.
Suppose $\mathcal{S}$ is closed under pairwise intersection. Let $A, B, C$ be given such that they are in $\mathcal{S}$. Then $A\cap B\in\mathcal{S}$. Since $C\in\mathcal{S}$, $(A\cap B)\cap C\in\mathcal{S}$. Induction gives us that any finite intersection is also in $\mathcal{S}$.
So it seems like they are really the same concept, right?

The operations are not equivalent, because closure under empty intersections does not follow from closure under pairwise intersections. However, an empty intersection is clearly a finite intersection.
$\pi$-systems are families of sets which are closed under pairwise intersections (hence all non-empty finite intersections). However, they do not have to be closed under empty intersections, hence do not have to be closed under all finite intersections.
Your induction proof only works to show closure under non-empty finite intersections. In particular, it's not possible to induct "backwards" from $n=1,2$ to $n=0$ (empty intersection).
The question and answers on this page explain empty intersection better than I can.
See also this question.
