 Solution

Put $K={\bigcap }_i{F}_i$. If some $F_i$ is empty, so is $K$, which is connected. Otherwise the compact sets are closed in the Hausdorff ambient space, and can all be considered as nested nonempty closed subsets of the compact space $F_0$. Part (a)(ii) gives $K\ne \varnothing$, and $K$ is compact.

Suppose $K$ were disconnected. Part (c)(iii) gives disjoint ambient open sets $U,V$ covering $K$, both meeting it. Part (a)(iii), applied within $F_0$ to $U\cup V$, gives $F_N\subseteq U\cup V$ for some $N$. Both $U$ and $V$ meet $F_N$ because they meet $K\subseteq F_N$. Their intersections with $F_N$ would therefore disconnect $F_N$, a contradiction. Thus