Find a Quotient of Hausdorff spaces not is Hausdorff Give a example of a quotient of a Hausdorff space that is not Hausdorff.
Attempt
Let $X=\mathbb{R}$, for $x,y\in X$ we define $x\sim y$ if $x=y$ or $|x|=|y|>1$.
I claim that the quotient $X\setminus \sim$ not is Hausdorff.
Let $\pi: X \longrightarrow  X\setminus \sim$ given by $\pi(x)=[x]$ and define the topology on $X \setminus \sim$ given by $U\subset X\setminus \sim$ is open if $\pi^{-1}([U])\subset X$ is open .
For $x\in X$ we have two possible cases.
Case I
$x\in \mathbb{Q}$, namely $x=p/q$ where $(p,q)=1,q\neq 0,p,q\in \mathbb{Z}$ WLOG $|x|>1$and then $[x]=\lbrace \frac{r}{s}\in \mathbb{Q} \mid ps=qr\rbrace$$\bigcup$ $ \lbrace -\frac{r}{s}\in \mathbb{Q} \mid ps=qr\rbrace$.
Other wise $|x|<1$ then $[x]=\lbrace \frac{r}{s}\in \mathbb{Q} \mid ps=qr\rbrace$
Case II
$x\in \mathbb{I}$ then $[x]=\lbrace x,-x\rbrace$.
We want take use the following:
If X is Hausdorff and $x_i\in X$ for $i\in 1,2,3, \cdots n$  then $\lbrace x_1,x_2,x_3, \cdots x_n\rbrace$ is closed
Consider $[\frac{1}{n}]\in X \setminus \sim $ for some $n\in \mathbb{N}$ fixed.
$\pi^{-1}(\lbrace[\frac{1}{n}]\rbrace)=\lbrace \frac{k}{kn}\mid k\in \mathbb{N} \rbrace=\lbrace \frac{1}{n}, \frac{1}{n}, \cdots \frac{1}{n} \rbrace =\lbrace \frac{1}{n}\rbrace\subset X$ wich is closed.
Hence $\lbrace[\frac{1}{n}]\rbrace$ is closed in $X\setminus \sim$.
Now consider $\bigcup_{n\in \mathbb{N}}\lbrace \left[\frac{1}{n}\right]\rbrace  \in X\setminus \sim $ and notice that
$$\pi^{-1}\left(\bigcup_{n\in \mathbb{N}}\lbrace \left[\frac{1}{n}\right] \rbrace \right)=\bigcup_{n\in \mathbb{N}}\pi^{-1}\left( \lbrace \left[\frac{1}{n}\right] \rbrace \right)=\lbrace\frac{1}{n}\mid n\in \mathbb{N} \rbrace \subset \mathbb{R} $$
Which is clearly not closed since $0$ is limit point of the set and is not contained in it.
Hence $X \setminus \sim$ cannot be Hausdorff.
I miss some?
Is possible get a intuitive representation of this quotient?
 A: The example works, but there is no reason to consider rational and irrational numbers separately, and your argument to show that $X/\!\sim$ is not Hausdorff is incorrect. I will first give a correct argument and then explain what is wrong with yours.
Let $p=\pi(-1)$ and $q=\pi(1)$; $|-1|=|1|=1\not>1$, so $p\ne q$. Suppose that $U$ and $V$ are disjoint open nbhds of $p$ and $q$ respectively. Then $\pi^{-1}[U]$ is an open nbhd of $-1$ in $X$, so there is a positive $\epsilon_p<1$ such that $(-1-\epsilon_p,-1+\epsilon_p)\subseteq\pi^{-1}[U]$. Similarly, there is a positive $\epsilon_q<1$ such that $(1-\epsilon_q,1+\epsilon_q)\subseteq\pi^{-1}[V]$. Let $\epsilon=\min\{\epsilon_p,\epsilon_q\}$, and let $x=1+\frac{\epsilon}2$. Then $|x|>1$, so $\pi(x)=\pi(-x)$, and therefore
$$\begin{align*}
\pi(x)&\in\pi[(1-\epsilon,1+\epsilon)]\cap\pi[(-1-\epsilon,-1+\epsilon)]\\
&\subseteq\pi\big[\pi^{-1}[U]\big]\cap\pi\big[\pi^{-1}[V]\big]\\
&=U\cap V\,,
\end{align*}$$
so that $U$ and $V$ are not disjoint. $U$ and $V$ were arbitrary nbhds of $p$ and $q$, so $p$ and $q$ do not have disjoint nbhds, and the quotient is not Hausdorff.
As for your argument, it is true that all finite subsets of a Hausdorff space are closed, so finding a finite set that is not closed would indeed show that the space is not Hausdorff, but the set $\bigcup_{n\in\Bbb Z^+}\left\{\left[\frac1n\right]\right\}$ is not finite, so showing that it is not closed tells you nothing. (You could even note that it’s also not closed in $\Bbb R$, which certainly is Hausdorff.) In fact a space is $T_1$ if and only if all of its finite subsets are closed, and $X/\!\sim$ is $T_1$.
There is a very easy way to picture the quotient. Bend the closed interval $[-1,1]$ around into a circle, but don’t close it: leave $-1$ and $1$ as separate points. Glue the rays $(1,\to)=\{x\in\Bbb R:x>1\}$ and $(\leftarrow,-1)=\{x\in\Bbb R:x<-1\}$ together, attaching $x$ to $-x$ for each $x>1$. You should end up with a picture something like this (but better!):
          _
         / \ 1
        |   :------------------------------ 
         \_/ -1

Except for $1$ and $-1$, a point in the circular part has an open nbhd base of small open intervals. A point in the ray that is formed by gluing together $(1,\to)$ and $(\leftarrow,-1)$ also has an open nbhd base of small open intervals. A basic open nbhd of $1$ looks like an open interval that starts on the circle a little above $1$ and runs through $1$ into the ray. Similarly, a basic open nbhd of $-1$ looks like an open interval that starts on the circle a little below $-1$ and runs through $1$ into the ray. And such intervals necessarily intersect on the ray. (I apologize for the very shoddy sketch, but at the moment I do not have the ability to produce a proper diagram.)
