Proof of homeomorphism. Let $X = [0, 1] × \mathbb{R}.$
An equivalence relation is given by $(0, x) ∼ (1, x).$
Show that $X/∼$ is homeomorphic to $S^1 × \mathbb{R}$.
This is obviously true intuitively but I am not quite sure how to prove it.
I think that I can show that $[0,1]/∼$ is homeomorphic to $S^1$ by defining
$$f:[0,1]\rightarrow S^1: f(x)=(cos(2\pi x), sin(2\pi x)).$$ 
Since the following hold:
$(1)\ f(x)=f(y)\leftrightarrow x∼y$
$(2)\ f$ is onto
$(3)$ If $U$ is open in $[0,1]$, then $f(U)$ is open in $S^1$
$(4) f$ is continuous
there is a unique map
$$g:[0,1]/∼\rightarrow S^1$$
that is a homeomorphism.
Is this ok?
If so, is it ok to define $$f:[0,1]×\mathbb{R}\rightarrow S^1×\mathbb{R}: f(x,y)=(cos(2\pi x), sin(2\pi x),y)$$ 
and use the above argument?
Any feedback would be greatly appreciated.
 A: $(3)$ is false: $\left[0,\frac12\right)$ is open in $[0,1]$, but its image under $f$ is not open in $S^1$. 
However, if $\pi:[0,1]\to[0,1]/\{0,1\}$ is the quotient map, and you define $g:[0,1]/\{0,1\}\to S^1$ so that $g\circ\pi=f$ (which makes sense, since $f$ is constant on equivalence classes), the map $g$ will be a homeomorphism. Once you’ve worked out the details for that, use the same idea on a larger scale, replacing $f$ by $f\times\mathrm{id}_{\Bbb R}$ and $\pi$ by the appropriate quotient map.
A: Unfortunately, you cannot use the argument again, since $f\times Id_Y:X\times Y\to Z\times Y$ need not be a quotient map if $f:X\to Z$ is a quotient map. But it is true if $Y$ is locally compact, so in your case $f\times Id_{\mathbb R}$ is indeed a quotient map.
Also, if $f$ is open, then $f\times Id_Y$ is open for any $Y$ since the product of open maps is open. But your $f$ is not open, it is closed, being a continuous map from a compact to a Hausdorff space.
Edit: You can also show directly that $g:=f\times Id_{\mathbb R}$ is closed. There are many ways to show that a map is closed. One could say that closedness is local in the codomain, meaning that $h:X\to Z$ is closed iff each point $z\in Z$ has a neighborhood $V$ such that the restriction $\hat h:h^{-1}(V)\to V$ is closed. To apply this to $g$ consider the cover of $S^1\times\mathbb R$ via {$V_n=S^1\times[n,n+2]\mid n\in\mathbb Z\}$ and show that $\hat g:g^{-1}(V_n)\to V_n$ is closed.
