$X$ 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 $U$ be an open set containing $A=\cap_n A_n$. Then there exists $n_0$ such that $A_n \subset U$ for $n\ge n_0$ ( otherwise the sets $A_n \backslash U$ would be closed, non void and decreasing so their intersection would be nonvoid). 
Assume now $\cap_n A_n$ is not connected, $A=A'\cup A''$. Take $U'\supset A'$, $U''\supset A''$, open, disjoint. There exists $n_0$ such that $A_n\subset U'\cup U''$ for $n\ge n_0$. We conclude that $A_n$ is disconnected for $n\ge n_0$. 
