Define the following equivalence relation on $\mathbb{R}^2$: $(x,y)\sim(x',y')$ iff there is $n\in \mathbb{Z}:(x',y')=(x+n,(-1)^ny)$. Let be $E=\mathbb{R}^2/\sim$ the quotient space, and $q:\mathbb{R}^2\to E$ the quotient map.
Let $S:=[0,1]\times \mathbb{R}\subseteq\mathbb{R}^2$. Let $p:S\to E$ be the restricion of $q$ to $S$.
I want to show that $p$ is surjective and closed.
Surjectivity of $p$
Let $(x,y)\in \mathbb{R}^2$. I want to show there is $(x',y')\in S$ such that $(x,y)\sim(x',y')$. Let $n \in Z$ such that $n\leq x <n+1$, then $x-n\in [0,1]$, so I take $(x',y')=(x-n,(-1)^ny)$.
Closedness of $p$
Let $C$ be a closed subset of $S$. I think we have $p^{-1}(p(C))=C \cup \{(1,-y):(0,y)\in C\}\cup \{(0,-y):(1,y)\in C\}$. If this is true, it remain to prove that $A:=\{(1,-y):(0,y)\in C \}$ and $B:=\{(0,-y):(1,y)\in C\}$ are closed in $\mathbb{R}^2$.
Let $\tau:\mathbb{R}^2\to \mathbb{R}^2, (x,y)\to (x+1,-y)$. Then $\tau$ is an homeomorphism and $A=\tau(C)$ is closed in $\mathbb{R}^2$. Analogously $B=\tau^{-1}(C)$.
The first question is if my proof is correct, and the second question is what are other smarter/shorter/ more elegant proofs. Thank you in advance :)
Birth of the question: this is part of Example 10.3 in John Lee's Book Introduction to Smooth Manifolds
EDIT
My proof is wrong: to show that $p(C)$ is closed in $E$, I must show that $q^{-1}(p(C))$ is closed in $\mathbb{R}^2$, not that $p^{-1}(p(C))$ is closed in $[0,1]\times \mathbb{R}$.
So the question now is: how can I prove that $p$ is closed?
Since $S$ is closed in $\Bbb R^2$, $C$ is closed in $\Bbb R^2$. A set $C’=\{(x,-y):(x,y)\in C\}$ is closed in $\Bbb R^2$ too, thus the set
$$q^{-1}(p(C))=\bigcup_{k\in Z} ((2k,0)+C)\cup ((2k+1,0)+C’)$$
is closed in $\Bbb R^2$ as a union of a locally finite family of closed sets.