Sierpiński's Theorem/Lemma 1

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

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

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

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

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

Proof
Let $p \in C$.

By Quasicomponents and Components are Equal in Compact Hausdorff Space and Quasicomponent is Intersection of Clopen Sets, $C$ is the intersection of the set $\mathcal K$ of all subsets of $A$ containing $p$ that are clopen relative to $A$.

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

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

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

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

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

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

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 $X$.

Therefore $K$ is open in $X$.

Thus $K$ is clopen in $X$.

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

Therefore $X$ is not connected.

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