Component of Point is not always Intersection of its Clopen Sets

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

Let $x \in S$.

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

Let $K_x = \displaystyle \bigcap_{x \mathop \in K} K$ clopen in $T$.

Then it is not always the case that $\operatorname{Comp}_x \left({T}\right) = K_x$

Proof
Note that from Clopen Set contains Components of All its Points:
 * $\operatorname{Comp}_x \left({T}\right) \subseteq K_x$

It remains to be demonstrated that it is not always the case that $K_x \subseteq \operatorname{Comp}_x \left({T}\right)$.

Let $T$ be the nested rectangle space in the Euclidean plane.

Let $L_1$ and $L_2$ be the boundary lines of $T$.

Let $x \in L_1$.

From Boundary Line in Nested Rectangle Space is Component $L_1$ is a component of $T$.

That is:
 * $L_1 = \operatorname{Comp}_x \left({T}\right)$

From Union of Boundary Lines in Nested Rectangle Space is Quasicomponent, $L_1 \cup L_2$ is a quasicomponent of $T$.

By Quasicomponent is Intersection of Clopen Sets:
 * $L_1 \cup L_2 = \displaystyle \bigcap_{x \mathop \in K} K$ clopen in $T$

Thus, while $\operatorname{Comp}_x \left({T}\right) \subseteq L_1 \cup L_2$, it is not the case that $L_1 \cup L_2 \subseteq \operatorname{Comp}_x \left({T}\right)$.

Hence the result.