I'm aware how to prove that the intersection of nested sequence of compact non-empty sets is compact and non-empty. but I want to generalize this question to transfer the hypothesis of having nested sequence of compact non-empty sets with having a collection of closed subsets which have finite intersection property. so, the new question is:
suppose that $M$ is covering compact and $\mathcal{C}$ is a collection of closed subset of $M$ such that every intersection of finitely many members of $\mathcal{C}$ is non-empty. Prove that the grand intersection $\bigcap _{A_n \in \mathcal{C}} A_n$ is non-empty.
In first case assume that the elements of nested sequence are $B_1 \supset B_2 \supset B_3 ... \supset B_n ...$ by compactness of each element of nested sequence we can say that the intersection is compact. then because each $B_n$ is non-empty, so for each $n \in \mathbb{N}$ we can choose a member $b_n \in B_n$ and make the sequence $(b_n)$. $(b_n) \in B_1$ Since the each $B_n$ is contained in $B_1$. Since $B_1$ is compact, $(b_n)$ has a convergent subsequence in $B_1$ named $ (b_{n_k})$ which converges to a point $p \in B_1$. the limit $p$ also lies in the set $B_2$ since except possibly for the first term, the subsequence $(b_{n_k})$ lies in $B_2$ and $B_2$ is a closed set. The same is true for each $B_n$, so $ p \in \bigcap{B_N}$ and $\bigcap B_n$ is non-empty.
But in my question, we can just say the $\bigcap A_n$ is compact since each $A_n$ is compcat. But what about being non-empty?
