The boundary of a closed subset $A$ of a compact connected hausdorff space Let $X$ be a compact connected hausdorff space. Let $A\subseteq X$ such that $A$ is closed and $A\neq\emptyset,X$. Let $C$ be a (connected) component of $A$. Is it true that $C\cap\partial A\neq\emptyset$?
 A: Lemma 1: Connected components of any topological space are always closed sets in that space.
Proof: let $D$ be a connected component of some arbitrary topological space $Y$. Since closures of connected sets are connected, it follows that $\overline{D}$ is a connected set intersecting $D$. By definition of connected component, any connected set intersecting $D$ is contained in $D$, so it follows that $\overline{D} \subseteq D$. Thus $\overline{D}=D$, so $D$ is closed). $\Box$
Lemma 2: Let $X$ be a compact Hausdorff space, and let $U \subset X$ be open. If $C$ is a compact connected component of $U$, then $C$ is a connected component of $X$.
Proof: (This is the hardest part.) Apply the second lemma in this answer, with $Y=U$ and $Z=C\cup(X \backslash U)$. $\Box$
Returning to the original question, we see by Lemma $1$ that $C$ is a closed (hence compact) subset of $X$.
Assume for contradiction that $C \cap \partial A = \emptyset$. Since $C \subset A$, this means that $C$ and $\overline{X \backslash A}$ are disjoint closed subsets of the normal space $X$. Thus there exist disjoint open sets $U,V$ such that $C \subset U$ and $\overline{X \backslash A} \subset V$. Since $C$ is a connected component of $A$ and since $U \subset A$, it follows that $C$ is a connected component of $U$. By Lemma $2$, we see that $C$ is also a connected component of $X$. By connectedness of $X$ this implies that $C=X$ or $C=\emptyset$, contradiction.
