Clopen Set contains Components of All its Points

Theorem
Let $T = \left({X, \tau}\right)$ be a topological space.

Let $S \subseteq X$ be both closed and open in $T$.

Then $S$ contains the components of all of its points.

Proof
Let $S$ be a clopen set in $T$.

By definition, $S$ is open and so $S \in \tau$.

But as $S$ is also closed, by definition $\complement_X \left({S}\right) \in \tau$ where $\complement_X$ denotes complement relative to $X$.

Thus $S$ and $\complement_X \left({S}\right)$ are both open such that:
 * $S \cap \complement_X \left({S}\right) = \varnothing$ from Intersection with Relative Complement is Empty
 * $S \cup \complement_X \left({S}\right) = X$ from Union with Relative Complement

and so forming a partition of $T$.

As $S$ and $\complement_X \left({S}\right)$ are both closed, it follows from Closed Set Equals its Closure that:
 * $S \cap \left({\complement_X \left({S}\right)}\right)^- = \varnothing = S^- \cap \complement_X \left({S}\right)$

and so by definition $S$ and $\complement_X \left({S}\right)$ are separated.

Suppose the hypothesis were false, and that:
 * $\exists x \in S: \operatorname{Comp}_x \left({T}\right) \nsubseteq S$

where $\operatorname{Comp}_x \left({T}\right)$ is the component of $x$ in $T$.

Then:
 * $\exists z \in \operatorname{Comp}_x \left({T}\right): z \in S \cup \complement_X \left({S}\right)$

This means that $S \cup \complement_X \left({S}\right)$ is connected.

This contradicts the fact that $S$ and $\complement_X \left({S}\right)$ are separated.