Let $(B_i)_{i\in I}$ be an indexed family of closed, connected sets in a compact space X. Suppose $I$ is ordered, sucht that $i < j \implies B_i \supset B_j$.
Is $B = \bigcap_i B_i$ necessarily connected?
I can prove it, if I assume $X$ to be Hausdorff as well: If $B$ is not connected, then there are two disjoint, closed, nonempty sets $C$, $D$ in $B$, such that $C \cup D = B$. Now these sets are also closed in $X$, hence by normality there exist open disjoint neighborhoods $U$, $V$ of $C$ and $D$, respectively.
Then for all $i$: $B_i \cap U^c \cap V^c \ne \emptyset$, since $B$ is contained in $B_i$ and $B_i$ is connected. Thus we must also have
$ B \cap U^c \cap V^c = \bigcap_i B_i \cap U^c \cap V^c \ne \emptyset $
by compactness and the fact that the $B_i$ satisfy the finite intersection property. This is a contradiction to the choice of $U$ and $V$.
I can neither see a counterexample for the general case, nor a proof. Any hints would be greatly appreciated!
Thanks,
S. L.