$C\subseteq X$ is called a component of $X$ is $C$ is a maximal connected subset of $X$.
A theorem says, let $\sim$ be an equivalence relation defined as $x\sim y$ iff $x$ and $y$ are in the same component (it's easy to check that $\sim$ is an equivalence relation). Then the equivalence classes of $\sim$ are the components of $X$.
To show that the equivalence classes are precisely the components of $X$, I think it suffices to show if $C_1$ and $C_2$ are different components of $X$, then $C_1\cap C_2=\emptyset$.
So suppose for a contradiction $C_1\cap C_2\neq\emptyset$. Clearly one cannot be a subset of another, otherwise one of them can't be a component since it would violate maximality. So there exist some $x\in C_1\setminus C_2$ and $y\in C_2\setminus C_1$. How can I get a contradiction from this?
Best Answer
Haven't you arrived at the solution already. If $z \in C_1 \cap C_2$, then x~z and y~z. Therefore, y~x from the definition of the classes. So you have a contradiction as this shows that neither $C_1$ nor $C_2$ is maximally connected and are thus not components.