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 $p \in C$.
Aiming for a contradiction, suppose $C$ 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.
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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 $X \setminus \paren {U \cup V}$.
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Thus by compactness has a finite subcover $T'$.
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Let $T$ be the set of complements of elements of $T'$.
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Then $\ds C \subseteq \bigcap T \subseteq U \cup V$.
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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 $C \subseteq U' \cup V' = T$.
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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 $\struct {X, \tau}$.
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It follows that $C$ is connected.
$\blacksquare$