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.

A subset $A$ is connected when it is a connected space in its subspace topology: there is no separation of a topological spaceinto two nonempty disjoint open subsets of $A$. Equivalently, the only clopen sets of $A$ are $\varnothing$ and $A$.

Since the given $A$ is not connected, choose such $P,Q$. Each is the complement of the other in $A$, so both are also closed subsets of $A$. They are compact by the closed subset of a compact space result. We now separate these two compact subsets by ambient open sets.

For each $p\in P$, part (b), applied to $Q$ and $p$, gives disjoint ambient open sets $G_p,H_p$ with $p\in G_p$ and $Q\subseteq H_p$. Choose $p_1,\dots ,p_m$ such that the $G_{p_j}$ cover $P$, using its compactness. DefineThen $P\subseteq U$, $Q\subseteq V$, and $U,V$ are disjoint open sets. Indeed a point of $U$ lies in some $G_{p_j}$, whereas every point of $V$ lies in the corresponding $H_{p_j}$. ConsequentlyThis proves the requested ambient separation, using the more general separation of disjoint compact subsets. The initially given relatively open sets $P,Q$ need not themselves be open in $X$; the finite construction supplies ambient open neighbourhoods.

For every $k\in K$, the Hausdorff condition gives disjoint open neighbourhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ form an open cover of $K$, so choose a finite subcover $U_{k_1},\dots ,U_{k_m}$. PutThen $U,V$ are open sets, $K\subseteq U$ and $x\in V$. If a point belonged to $U\cap V$, it would belong to some $U_{k_j}$ and to $V_{k_j}$, contrary to their choice. ThusThe finiteness matters: a finite intersection of open sets is open. This is separation of a point and a compact set. If $K=\varnothing$, use $U=\varnothing$, $V=X$.

Every $x\in X\setminus K$ consequently has an open neighbourhood $V$ contained in $X\setminus K$. The complement is open, so a compact subset of a Hausdorff space is closed.

The conclusion can fail without the Hausdorff condition. Take the two-point set $X=\{0,1\}$ with the indiscrete topology $\{\varnothing ,X\}$. The singleton $K=\{0\}$ is compact, since one member of any open cover suffices. It is not closed, because its complement $\{1\}$ is not open. The two points cannot have disjoint open neighbourhoods, so this example is indeed not Hausdorff.