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Write $K={\bigcap }_{i\in I}{K}_i$. Each $K_i$ is closed in $X$, by the compact subset of a Hausdorff space theorem. Therefore $K$ is closed, and is compact because $X$ is compact. The nonempty finite intersections give the finite intersection property, so part (a)(ii) also yields $K\ne \varnothing$.

Suppose $K$ were not connected. Part (c) would give disjoint open sets $U,V$ containing $K$ in their union and each meeting $K$. LetThis is a closed subset of the compact space $X$, and $C\cap {\bigcap }_{i\in I}{K}_i=\varnothing$. Apply the contrapositive of the finite intersection characterization of compactness to the family of closed sets $\{C\}\cup \{K_i:i\in I\}$. Some finite subfamily has empty intersection. Since every finite intersection of just the $K_i$ is nonempty, that subfamily must include $C$. Thus for some finite $J\subseteq I$,But $K\subseteq K_J$ and $K$ meets both $U$ and $V$. Therefore $K_J\cap U$ and $K_J\cap V$ are nonempty disjoint relatively open sets covering $K_J$, contradicting its assumed connectedness. This provesThis is connectedness of an intersection from connected finite intersections. If $I=\varnothing$, its intersection is $X$; the stated hypothesis for the finite set $J=\varnothing$ already makes $X$ nonempty and connected, so that case is included.