Connected components of a space $X$ are disjoint $C\subseteq X$ 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?
 A: 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.
A: Basic fact: if $C,D$ are connected subsets of $X$ and $C \cap D\neq \emptyset$ then $C \cup D$ is connected.
Proof: suppose that $C \cup D = U \cup V$ where $U \cap V = \emptyset$ and $U,V$ are open in $C \cup D$. Let $p \in C \cap D$. This $p$ either lies in $U$ or $V$, say in $U$ for definiteness. 
Then $C= (U \cap C) \cup (V \cap C)$ and both sets $U \cap C$ and $V \cap C$ are open in $C$, and they are still disjoint.
As $C$ is connected, this cannot be a real disconnection of $C$, so one set is empty, the other the whole set. $U \cap C$ contains $p$, so that is the non-empty one. So $U \cap C = C$, which means that $C \subseteq U$.
Then also $D = (U \cap D) \cup (V \cap D)$ which for the same reason cannot be a disconnection of $D$, and again $p \in U \cap D$, so $U \cap D = D$ and $D \subseteq V$. Hence $C \cup D = U $ and so $V = \emptyset$. This shows that $C \cup D$ is connected.
Now, if $C_1$ and $C_2$ are two components, if they intersect, then $C_1 \cup C_2$ is connected by the previous fact. And as $C_1 \subseteq C_1 \cup C_2$ and $C_1$ is a maximal connected set, $C_1 = C_1 \cup C_2 = C_2$ (the last by maximality of $C_2$. But this means that $C_1 = C_2$. So if components intersect they are equal, which is another way of saying different components are disjoint.
