Statement of the problem: Prove that if $A_1 \supseteq A_2 \supseteq A_3 \cdots$ are all finite, nonempty sets of real numbers, then the intersection $\bigcap_{n=1}^{\infty} A_n$ is also finite and nonempty.
My solution: $\bigcap_{n=1}^{\infty} A_n = \emptyset.$ Then at least one $A_n$ is disjoint from the rest - contradiction, since each $A_n \subseteq A_{n-1}.$ Next assume that the intersection has infinite cardinality - contradiction, since $A_n \supseteq \bigcap_{n=1}^{\infty} A_n,$ but we have $A_n$ finite.
This is a homework problem and this is what I will be turning in regardless of any answers I get. I would like to know if there is a direct proof of this claim. I had considered trying to show that $\exists x \in A_n$ for each choice of $n,$ which would show that the intersection is nonempty, but:a) I don't know how to do this, and b) I still would not be able to prove the intersection was finite except by contradiction.
Thanks!
edit: Moreover, we were told not to use induction on this problem. But if $A_1$ is finite, can't we reduce the collection of $A_n's$ to a subset of the power set of $A_1$ and then induct?