Equivalence of Definitions of Component

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

Let $x \in T$.

Proof
Let $C = \operatorname{Comp}_x \left({T}\right)$.

$(1) \iff (2)$
By definition, $y \in C$ there exists a connected set of $T$ that contains both $x$ and $y$.

$(2) \iff (3)$
From $(2)$ and Space with Connected Intersection has Connected Union, $C$ is a connected set of $T$.

Let $A$ be a connected set of $T$ that contains $x$.

Then by $(2)$ we have $A \subseteq C$.