Closed Subset of Connected Space with Boundary a Single Point is Connected Old qual question here.

Let $X$ be connected and $C\subset X$ be closed such that the boundary of $C$ is a point. Show that $C$ is connected.

Attempt:
For contradiction, suppose $U,V$ are nonempty disjoint sets open in $C$ such that $C=U\cup V$. Call $y=\partial C$ and say $y\in U$. Then $V\subset\operatorname{int}C$, which is open, so $V$ is open in $X$. Since $C$ is closed in $X$, $C^C$ is open in $X$.
This is where I'm stuck. I think that $U\cup C^C$ has to be open in $X$, then $X=(U\cup C^C)\cup V$ would not be connected, a contradiction. But I'm stuck here.
 A: Note that since $C$ is closed, then $\partial C = C \setminus \operatorname{Int} (C)$, so $C = \{ x \} \cup \operatorname{Int}(C)$ where $\partial C = \{ x \}$.
Suppose that $C$ is not connected, so that there are $U,V$ open subsets of $X$ such that


*

*$U \cap C \neq \emptyset \neq V \cap C$,

*$C \subseteq U \cup V$,

*$(U \cap V ) \cap C = \emptyset$.


Without loss of generality, $x \in U$. Then as $x \notin V$ by replacing $V$ with $V \cap C = V \cap \operatorname{Int} ( C )$ we may assume that $V \subseteq C$, and so in particular $V \cap U = \emptyset$.
Consider $U \cup ( X \setminus C )$, and $V$.


*

*Clearly $U \cup ( X \setminus C)$ and $V$ are open subsets of $X$.

*Note that $( U \cup ( X \setminus C ) ) \cup V = ( U \cup V ) \cup ( X \setminus C ) \supseteq C \cup ( X \setminus C ) = X$.

*Note that $( U \cap ( X \setminus C ) ) \cap V = ( U \cap V ) \cup ( ( X \setminus C ) \cap V )  = \emptyset$.


Therefore $U \cup ( X \setminus C )$ and $V$ form a separation of $X$, contradicting that $X$ is connected! Thus it must be that $C$ is connected.
A: Here’s a slightly simpler argument (along the same general lines).
Suppose that $p$ is the unique boundary point of $C$, and note that $\operatorname{int}C=C\setminus\{p\}$. If $C$ is not connected, there are non-empty sets $H$ and $K$, both relatively closed in $C$, such that $H\cap K=\varnothing$ and $H\cup K=C$. $C$ is closed, so in fact $H$ and $K$ are closed in $X$. Without loss of generality assume that $p\in K$. Then $H=(\operatorname{int}C)\setminus K$, which is open, so $H$ is clopen (i.e., both closed and open). Since $\varnothing\ne H\ne X$, it follows immediately that $X$ is not connected: $H$ and $X\setminus H$ form a separation of $X$.
(If you don’t immediately see why $(\operatorname{int}C)\setminus K$ is open, note that it equals $(\operatorname{int}C)\cap(X\setminus K)$, the intersection of two open sets in $X$.)
A: Let $\partial C=\{p\}$, and let $V,W$ be open subsets of $X$ such that $C\subseteq V\cup W$ and $C\cap V\cap W=\emptyset$. We assume WLOG that $p\in V$, and our objective is to show that $B=\emptyset$, where $B:=C\cap W$.
If $A:=C^c\cup V$ then $A$ is open and nonempty ($p\in A$), $X=A\cup B$ and $A\cap B=\emptyset$, so by connectedness of $X$ it is suffices to show that $B$ is open. Now from $C=\operatorname{int}(C)\cup\{p\}, p\in V$ and $C\cap V\cap W=\emptyset\,$ we obtain $B\subseteq\operatorname{int}(C)$, so $B=\operatorname{int}(C)\cap W$, as desired.
