$X$ a compact Hausdorff space and a sequence $A_1 \supset A_2 \supset \dots$ of closed connected subsets. Show that $\bigcap A_n$ is also connected. X a compact Hausdorff space and a sequence $A_1 \supset A_2 \supset \dots$ of closed connected subsets.
I need to show that the intersection $\bigcap _{n\in\Bbb N}A_n$ is also connected.
I know that from the compactness of $X$ that the intersection is non-empty by the F.I.P criterion. if I choose a converge sequence of points $(x_n)$ so for every $n\in \Bbb N, x_n\in A_n$ we have a single limit $x \in \bigcap_{n\in\Bbb N}A_n$ (by the fact that X is Hausdorff), but I don't know how to show that $\bigcap _{n\in\Bbb N}A_n$ is connected...

Let $A = \bigcap_n A_n$. You already know that $A$ is non-empty and compact (intersection of compact sets in a Hausdorff space). We also know that $X$ is normal, being compact and Hausdorff.
Suppose $A$ were not connected. Then $A = C_1 \cup C_2$ where $C_1, C_2$ are disjoint non-empty closed subsets of $A$ (and hence also closed in $X$).
By normality, we have open disjoint subsets $C_1 \subseteq O_1$ and 
$C_2 \subseteq O_2$. Let $O = O_1 \cup O_2$.
So we know that $\bigcap_n A_n \subseteq O$. 
Claim: there exists some $N$ such that $A_N \subseteq O$.
Proof: suppose not. Then all sets $B_n = A_n \setminus O$ would be non-empty, compact (as $B_n$ is also closed), decreasing (as the $A_n$ are), and so there would be some $p \in \bigcap B_n$ (F.I.P. criterion again), which would be a point of $A$ that is not in $O$, contradiction. So such an $N$ must exist.
But then the $O_1, O_2$ disconnect $A_N$ (as both $O_i \cap A_N \supseteq C_i \neq \emptyset$, etc.) and this contradicts the connectedness of all $A_n$, which was a given.
So $A$ is connected.
