Quasicomponent of Compact Hausdorff Space is Connected

Theorem
Let $\struct {X, \tau}$ be a compact Hausdorff space.

Let $C$ be a quasicomponent of $\struct {X, \tau}$.

Then $C$ is connected.

Proof
Let $p \in C$.

$C$ is not connected.

Therefore, by definition of connected, there exist disjoint closed sets $A, B$ of $\struct {X, \tau}$ such that $C = A \cup B$.

By Compact Hausdorff Space is T4, there exist disjoint open sets $U, V$ of $\struct {X, \tau}$ such that $U \supseteq A$ and $V \supseteq B$.

By Quasicomponent is Intersection of Clopen Sets, $C$ is the intersection of all clopen sets of $\struct {X, \tau}$ containing $p$.

Since $U$ and $V$ are open, $X \setminus \paren {U \cup V}$ is closed.

Hence $X \setminus \paren {U \cup V}$ is compact.

Let $S$ be the set of clopen sets of $\struct {X, \tau}$ containing $p$.

Let $S'$ be the set of complements relative to $\struct {X, \tau}$ of elements of $S$.

Then $S'$ is an open cover of $X \setminus \paren {U \cup V}$.

Thus by compactness has a finite subcover $T'$.

Let $T$ be the set of complements of elements of $T'$.

Then $\ds C \subseteq \bigcap T \subseteq U \cup V$.

Furthermore, since $T$ is a finite set of clopen sets of $\struct {X, \tau}$]], $\bigcap T$ is clopen.

Let $\ds U' = U \cap \bigcap T$ and let $\ds V' = V \cap \bigcap T$.

Then $C \subseteq U' \cup V' = T$.

Since $T$ is clopen, so is $U'$.

But $C$ contains points in $U'$ and points not in $U'$, contradicting the fact that $C$ is a quasicomponent of $\struct {X, \tau}$.

It follows that $C$ is connected.