a) Let X be a topological space, and $\mathcal{A}$ a family of closed, compact subspaces. Then, if $\bigcap_{A \in \mathcal{A}}A \subseteq U$, and $U$ is open, then $\exists\ \mathcal{F} \subseteq \mathcal{A}$ such that $\mathcal{F}$ is finite and $\bigcap_{A \in \mathcal{F}}A \subseteq U$
b) Let X be a topological Hausdorff ("$T_2$" for some people) space, and $\mathcal{A}$ a family of compact subspaces such that $\bigcap_{A \in \mathcal{F}}A$ is connected for every finite $\mathcal{F}$. Prove that $\bigcap_{A \in \mathcal{A}}A$ is connected.
Proof: a) Suppose the contrary, then $\bigcap_{A \in \mathcal{F}}A \cap U^c \neq \emptyset$. By the Finite intersection Property, ($\bigcap_{A \in \mathcal{A}}A \cap U^c$ is compact), we arrive at a contradiction.
b) Here's where I get stuck... Suppose the contrary.
First, an observation: as $X$ is Hausdorff, so the subspaces are also closed.
Then, $\exists\ U, V$ open (there) such that $B:= \bigcap_{A \in \mathcal{A}}A = U \cup V$, and $U$ and $V$ are disjoint.
Then, $\exists\ U', V'$ open in $X$ such that $U = B \cap U'$, $V = B \cap V'$.
From there, I can get a finite $F$ such that $C:= \bigcap_{A \in \mathcal{F}}A \subset U' \cup V'$. I'd like to shrink the size of U' and V' so I disconnect C, but I couldn't so far. I think I haven't used "enough" the fact that X is Hausdorff.
Can anyone help me? Thanks!
Pick some $A_i\in \mathcal A$, suppose $\mathcal A=\{A_j:j\in J\}$. Then $A_i\subseteq U\cup \bigcup_{i\neq j} A_j$: if $x\in A_i$ is in every $A_j$; then $x\in U$, if there is some $j$ such that $x\notin A_j$; then $x\in A_j^c$. The $A_j$ are assumed to be closed so this open cover of $A_i$ must admit a finite subcover and we can assume it contains $U$, say $A_i\subseteq U\cup A_1^c\cdots \cup A_n^c$ (I've relabeled the $A_j$ in the finite subcover for ease of notation). Then evidently $\bigcap_{i=0}^n A_i\subseteq U$, where $A_0:= A_i$.