Let us agree that a map is a continuous function.
A quotient map $q : X \to Y$ is a surjective map such that $V \subset Y$ is open if and only $q^{-1}(V) \subset X$ is open.
A quotient map is in general not an open map, but is is well-known that any open surjective map is a quotient map (this follows from $q(q^{-1}(V)) = V$).
The assumption "Let $q : X \to Y$ be an open quotient map" is therefore the same as "Let $q : X \to Y$ be an open surjective map".
What is the relation to the post $X/{\sim}$ is Hausdorff if and only if $\sim$ is closed in $X \times X$?
In this post we do not start with a map $q : X \to Y$ between topological spaces, but with a space $X$ and an equivalence relation $\sim$ on $X$ and then define $Y = X / \sim$. The function
$$q : X \to Y, q(x) = [x] ,$$
where $[x]$ denotes the equivalence class of $x$ with respect to $\sim$, is by definition a surjection. $X$ is a topological space, but $Y$ is defined as a set which does not yet have a topology. In this situation the set $Y$ is by default endowed with the quotient topology (which is the finest topology on $Y$ making $q$ continuous).
Thus, the phrase "If the quotient map is open" tells us two facts:
(1) The set $Y$ is endowed with the quotient topology.
(2) $q$ is an open map.
For the proof we only need (2). But without specifying the topology on $Y$ it does not make sense to say that $q : X \to Y$ is an open map, and that is the reason why the above phrase is used.
However, alternatively one could also say that $Y$ is endowed with a topology such that $q$ becomes an open map. In that case the topology on $Y$ must automatically be the quotient topology.
In your post we already have a topology on $Y$, and it is redundant to say that $q$ is an open quotient map.
Finally, the product of two open quotient maps is an open quotient map simply because the product of two open maps is an open map.
Best Answer
The equivalence relation says $(x,y)\sim(x',y')$ iff $x'-x\in\Bbb{Z}$ and $y'=\pm y$ which implies $E=[0,1)\times(\Bbb{R}^+\cup\{0\})$.
Denote the restricted domain by $D=[0,1]\times\Bbb{R}$, then consider the restricted map $q^*:D\to E$ defined explicitly by,
$$ q^*(x,y)= \begin{cases} (x,|y|) &\text{ if }x\in[0,1)\\ (0,|y|) &\text{ if }x=1\\ \end{cases} $$
The coordinate function for $y$ is obviously closed.
Take a closed set $U\subset D$ then the $x$-coordinate of elements in $U$ must take values in $[a,b]\subset[0,1]$, while the $x$-coordinate for the image must take values in $[a,b]$ OR $[a,1)\cup\{0\}$, both of which situations are closed since $[a,1)\cup\{0\}=([a,1]\cap [0,1))\cup(\{0\}\cap [0,1))$ (subspace). Because this is true for every closed set in $D$, so $q^*$ is closed.
Add one method (according to the note from hunter):
We can also prove $q:\Bbb{R}^2\to E$ is closed. Take a closed set $V\subset\Bbb{R}^2$ and denote the range for the $x$-coordinate of elements in $V$ by $V_x=[m,n]$. Then, $$q(V)_x= \begin{cases} [m_1,n_1]\subset[0,1) &\text{ if } \Bbb{Z}\cap[m,n]=\varnothing\\ [|m|-\lfloor|m|\rfloor,1)\cup[0,|n|-\lfloor|n|\rfloor] &\text{ if } \Bbb{Z}\cap[m,n]\neq\varnothing \end{cases} $$ So, $q$ is closed, and so does $q^*$ by restricting its domain, using the properties of a subspace.
General tips:
This is hard to generalize a tip for every situations, but the ideas are similar. I'm sure that professors have deeper understandings and more methods than I do.
To prove a map is open(closed), we can test the basis or subbasis elements (or closed sets/ complement of open sets). Sometimes it's easy, if we're given a homeomorphism, then it is immediately open and closed.
However, continuity doesn't imply anything about closedness or openness, e.g. $$ f(x)= \begin{cases} x, & x\ge0\\ 0, & x\le0 \end{cases} $$ $f$ is clearly continuous, but not open since $(-2,2)\mapsto[0,2)$ given the euclidean topology. Another one is $g(x)=|x|$, it is closed but not open since $(-1,2)\mapsto[0,2)$. Besides that, there exists continuous maps that are not open nor closed.
But the two explicit maps mentioned above are open when we're given the lower-limit topology. So,
I Hope this is useful.