Connected components Let $X$ be a topological space. We define $x\sim y$ if and only if $C_x=C_y$ (their connected components).
Let $X/\sim$ be the quotient space. 
I want to prove that $C_{\bar{x}}=\{\bar{x}\}$ for each $\bar{x}\in X/\sim$.
By contradiction, I suppose that $\{\bar{x},\bar{y}\}\subseteq C_{\bar{x}}$ for some $\bar{y}\neq\bar{x}$. This is $C_x\neq C_y$ in $X$. 
I don't know how to continue from here.
Any hint? Thanks.

Allow me to introduce the notation $\pi: X \to X/\sim$ for the projection map, so that $\overline x = \pi(x)$. I assume the definition you are familiar with for the quotient topology is that $U \subset X/\sim$ is open if and only if $\pi^{-1}(U)$ is open (if not, feel free to ask for clarification).
Suppose $C_{\overline x}$ contains two or more points. Then we get that $C_{\overline x}$ is a connected topological space, but $\pi^{-1}(C_{\overline x})$ is not. What can be said about a pair of open sets (in $X$) disconnecting $\pi^{-1}(C_{\overline x})$?
