Sierpiński's Theorem/Lemma 1

Theorem
Let $\left({S, \tau}\right)$ be a compact connected Hausdorff space.

Let $A$ be a closed, non-empty proper subset of $S$.

Let $C$ be a component of $A$.

Then:
 * $C \cap \partial A \ne \varnothing$

where $\partial A$ denotes the boundary of $A$.

Proof
Let $p \in C$.

Let $\mathcal V$ be the set of all subsets of $A$ containing $p$ that are clopen relative to $A$..

By Quasicomponents and Components are Equal in Compact Hausdorff Space and Quasicomponent is Intersection of Clopen Sets:
 * $C$ is the intersection of $\mathcal V$.

Aiming for a contradiction, suppose:
 * $C \cap \partial A = \varnothing$

By Boundary of Set is Closed, $K \cap \partial A$ is closed for each $K \in \mathcal V$.

Thus by Compact Space satisfies Finite Intersection Axiom, there exists a finite set $\mathcal V' \subseteq \mathcal V$ such that $\partial A \cap \bigcap \mathcal V' = \varnothing$.

But then:
 * $\displaystyle K = \bigcap \mathcal V' \in \mathcal V$

Therefore there exists a $K \in \mathcal V$ such that $K \cap \partial A = \varnothing$.

Since $A$ is closed in $S$, and $K$ is clopen in $A$, $K$ is closed in $S$.

We have that:
 * $\partial A = A^- \setminus \operatorname{Int} \left({A}\right)$

where $A^-$ is the closure of $A$ and $\operatorname{Int} \left({A}\right)$ is the interior of $A$

Hence as $K \subseteq A$, it follows that $K \subseteq \operatorname{Int} \left({A}\right)$.

Since $K$ is open relative to $A$, it is open relative to $\operatorname{Int} A$.

We have that $\operatorname{Int} \left({A}\right)$ is open in $S$.

Therefore $K$ is open in $S$.

Thus $K$ is clopen in $S$.

We have that $p \in K \subseteq A \subsetneqq S$.

Therefore $S$ is not connected.

From this contradiction it follows that:
 * $C \cap \partial A \ne \varnothing$