A quotient space of a Hausdorff space need not be Hausdorff The example is as follows:
$\mathbb{R}\times \{0,1\}$ is clearly a Hausdorff space. But the quotient space of $\mathbb{R}\times \{0,1\}$ defined by the equivalence relation $\langle x,0\rangle$~$\langle x,1\rangle \iff x\neq 0$ is not Hausdorff.
Why is that?
Is the quotient space the set $Q=\mathbb{R}$, such that $U\subset Q$ is open if and only if $U\times\{0\}\cup U\times\{1\}$ is open in $\mathbb{R}\times\{0,1\}$?
 A: 
Is the quotient space the set $Q=\mathbb{R}$, such that $U\subset Q$ is open if and only if $U\times\{0\}\cup U\times\{1\}$ is open in $\mathbb{R}\times\{0,1\}$?

No. The quotient space, by definition, is the set $X/_\sim$ where $X$ is the original space (in your case, $X=\mathbb R\times\{0,1\}$. This means $Q=\{[(x,y)]| (x,y)\in X\}$.
In $Q$, a set $U\subset Q$ is open if $q^{-1}(U)$ is open in $X$.

Now, for $x\neq 0$, you have $[(x,0)] = [(x,1)] = \{(x,0),(x,1)\}$, and you can certainly see that the local neighborhood around the point $[(x,0)]\in Q$ is homeomorphic to the local neighborhood around the point $x\in R$, but that does not mean that the two sets are equal! In fact, you get into trouble around $x=0$, since $[(0,0)] = \{(0,0)\}$ and $[(0,1)]=\{(0,1)\}$

So, the idea is to prove that those two points do not have disjoint neighborhoods.

Take any neighborhood $O_0$ containing $[(0,0)]$. Then, $q^{-1}(O)$ is an open set around $(0,0)$, which means that it contains some "interval" (i.e. there exists some $\epsilon_0 > 0$ such that $(-\epsilon_0, \epsilon_0)\times\{0\}\subset q^{-1}(O_0)$. 
From this, it should be easy to show that $I_0 = \{[(x, 0])| |x|<\epsilon_0\}$ must be a subset of $O_0$
If you now repeat the process on a neighborhood $O_1$ containing $[(0,1)]$, you will find that $I_1 = \{[(x,0)]; |x<\epsilon_1\}$ is a subset of $O_1$, and thus $O_0\cap O_1\neq\emptyset$
