not Hausdorff, question I am trying to understand an example for a space which is not Hausdorff.
I do not really see, why (Q1******) and (Q2******)marked underneath hold.
To show: $Y$ (see definition underneath) is not Hausdorff.
$X = [-1,1] \times \{0, 1 \} \subset \mathbb{R}^2$, with induced topology of $\mathbb{R}^2$ ($X$ is Hausdorff).
$Y = X /_{\sim}$, $\sim$ induced by: $(t,0) \sim (t,1) \ \forall t \in [-1,1] \setminus \{ 0 \}$
Let $P =(0,1), Q = (0,0)$. We will show: $P$ and $Q$ cannot be separated.
Let $U$ be open in $Y$, $P \in U$ and $V$ open in $Y$, $Q \in V$.
Let $p: X \to Y$ be the projection. Then
$p^{-1} (U) = U' \subset X$ open, $p^{-1} (V) = V' \subset X$ open.
Furthermore: $(0,1) \in U'$, $(0,0) \in V'$ and $U', V'$ open sets.
Obviously: $(- \epsilon, \epsilon) \times \{ 1 \} \subset U'$ since $(0,1) \in U'$.
$p( (-\epsilon, \epsilon)) \supset ((-\epsilon,0) \cup (0,\epsilon))
\cup \{ P \} $     (Q1******)
$\Rightarrow U' \supset ((- \epsilon, \epsilon) \times \{ 1 \} ) \cup
((- \epsilon, 0) \cup (0, \epsilon)) \times \{ 0 \} )$   (Q2******)
$\Rightarrow U' \supset ((- \epsilon, 0) \cup (0, \epsilon)) \times \{ 0,1 \} $
Similarly, $V' \supset ((- \epsilon', 0) \cup (0, \epsilon')) \times \{ 0,1 \} $
$\Rightarrow U' \cap V' \neq \emptyset$ 
$\Rightarrow U \cap V \neq \emptyset$.
 A: There's indeed a mild ambiguity in the notation.
To visualize $Y$, it's almost a single ordinary line segment, but its midpoint is taken as two different points (the two parallel line segments of $X$ are collapsed into one except for their midpoints).
   $$\_\_\_\_\_{}_:\_\_\_\_\_$$
Now, $(-\epsilon,0)\cup(0,\epsilon)$ in Q1 refers to a set of points on the two single line segment parts of $Y$ (each point with double cover in $X$), specifically to the set
$$\{[(a,0)]_\sim\,\mid\,-\epsilon<a<\epsilon,\,a\ne 0\}\ =\ \{[(a,1)]_\sim\,\mid\,-\epsilon<a<\epsilon,\,a\ne 0\}\,,$$
and indeed its preimage is $(-\epsilon,0)\times\{0,1\}\,\cup\,(0,\epsilon)\times\{0,1\}$.
If you add $P=[(0,1)]_\sim$ to this set, then it only adds the point $(0,1)$ in the preimage, as $P$ is not doubly covered.
The same goes for $Q$.
A: Notation: $f[S]=\{f(x):x\in S\}$ for any function $f$ and any $S\subset$ dom$(f).$
$<x,y>$ denotes an ordered pair. $(a,b), (a,b],[a,b)$ denote intervals in $\Bbb R.$
The quotient topology on $Y=p[X]$ is defined as the strongest topology such that $p$ is continuous.
The quotient map $p$ sends $x\in X$ to $p(x) =\{x'\in x:x'\sim x\}.$ 
So actually $p(Q)=\{x'\in X: x'\sim Q\}=\{Q\},$ and $p(P)=\{P\}.$ Also when $0<|x|\le 1$ and $j\in \{0,1\}$ we have $p(<x,j>)=\{<x,0>,<(x,1>\}.$ But this  is an inconveniently large amount of notation.
So define $p':Y\to Y'=[-1,0)\cup (0,1]\cup \{P,Q\}$ where $p'p(P)=P$ and $p'p(Q)=Q,$ and $p'p(<x,j>)=x$ when $0<|x|\le 1$ and $j\in \{0,1\},$ 
and let $S\subset Y'$ be open iff $(p')^{-1}S$ is open in $Y.$
Then $p':Y\to Y'$ is a homeomorphism, so $p(P),p(Q)$ cannot be separated in $Y$ iff their images $P, Q$ in $Y'$ cannot be separated in $Y'.$
For brevity let $p''=p'p.$
This use of $p'$ and $p''$ is common with quotient spaces, to give a simpler, more intuitive picture. 
Let $Q\in V$ where $V$ is open in $Y'.$ Then $(p'')^{-1}V$ is open  in $X$  so for some $e\in (0,1)$  we have $(p'')^{-1}V\supset (-e,e)\times \{0\}.$ Now  $p''$ is a surjection so $V=p''[(p'')^{-1}V].$ So we have $$V=p''[(p'')^{-1}V]\supset p''[(-e,e)\times \{0\}]=(-e,0)\cup (0,e)\cup \{Q\}.$$
Similarly if $P\in U$ where $U$ is open in $Y',$ then for some $e'\in (0,1)$  we have $$U=p''[(p'')^{-1}U]\supset p''[(-e',e')\times \{1\}]=(-e',0)\cup (0,e')\cup \{P\}.$$ Therefore $V\cap U\supset (0,\min(e,e'))\ne \emptyset.$
