Is the torus homeomorphic to this quotient space? I am working on this question:

Considering over $X=S^1\times\mathbb R$, with the usual induced topology  and the equivalence relation 
  $(u,t )R(u',t')$ if and only if $u'=\pm u, t-t' \in \mathbb Z$.
Tell if $(X/R,\epsilon_2/R)$  is T2, connected compact and homeomorphic to the torus $S^1\times S^1$

The equivalence relation tells me, if I am not wrong each equivalence class is formed of the symmetric pairs of points with respect to the orign in the xy plane and the corresponding points having z coordinate differing by an integer.
1)I am not very sure though, if that is what u=-u' over a circle means. It would make sense if it'd be (x,y)=-(x',y'), u could also represent an angle, and then my interpretation would be wrong. How should I interpret it?
2)Assuming it is correct, to prove it is T2, I need to show that for any pair of points of $X$ not in relation by $R$ there exist disjoint saturated open sets in $(X,\epsilon_2)$ separating the points. This is graphically understandable, but I am having trouble writing it down.
3) I need a function to prove the homeomorphism, would the usual parametrization of the torus do the trick?, that would  go from $[0,2\pi]\times [0,2\pi]$ to the torus, but still need to connect this to the given quotient. I've I can prove there is an homeomorphism, I can easily deduce compatness and connectedness
I've looked everywhere , but there is no info on this. Any help would be greatly appreciated
 A: Note that for the standard $S^1$, if $(x,y) \in S^1$, so is $(-x,-y)$.
So we can form an equivalence relation on $S^1$ where $u \sim u'$ iff $u=u'$ or $u=-u'$, so all classes are doubletons $\{(x,y), (-x,-y)\}\mid (x,y) \in S^1\}$. It is well known that $S^1{/}{\sim} \simeq S^1$ (the projective line is the circle) and the map that sends the class $\{u,u'\}$ to $u^2$ (as a complex number, so double the angle) is a homeomorphism.
This is what happens on te $S^1$ side of things. On the $\Bbb R$ side we have the standard modulo $\Bbb Z$ reduction, and $\Bbb R{/}\Bbb Z \simeq S^1$ should be well-known. So in the product we get that for your relation $R$ is the product of $\sim$ and the previous quotient (even holds as groups) and as $S^1$ and $\Bbb R$ are nice spaces (locally compact etc.) we get that $$(S^1 \times \Bbb R){/}R \simeq (S^1{/}\sim) \times (\Bbb R{/} \Bbb Z) \simeq S^1 \times S^1$$
which is what you want. A homeomorphism can be made explicit by mapping the class of $(u,t)=((\cos(t'), \sin(t')), t), t' \in [0,2\pi)$ to $(\cos(2t'), \sin(2t')),(\cos(2\pi t), \sin(2\pi t)) \in S^1 \times S^1$ e.g. 
