It is possible to deduce ($2$) from ($3$) without $q\times q$ being a quotient map.
Let $A\subset X$ be closed. Assume $R=\{(x,x')\in X\times X\mid q(x)=q(x')\}$ is closed, thus compact. Then $A\times X\cap R$ is compact, and so is also its image under the projection $p_2$ onto the second factor. But $p_2(A\times X\cap R)=\{x\in X\mid\exists a\in A:q(a)\sim q(x)\}=q^{-1}(q(A))$. So the saturation of a closed set is compact, and hence closed, which means that $q$ is a closed map.
One could also omit the compactness of $X$ if one assumes directly that $R$ is compact, because the last step only uses the Hausdorff'ness to deduce that a compact set is closed. Note that compactness of $R$ also makes $\{x\}\times X\cap R$ a compact set, so fibers are compact and this makes $q$ a so-called perfect map. These maps preserve many properties of the domain, for example all the separation axioms (except $T_0$)
This doesn't answer the question in the title, which I would really like to know myself. I only know about the product of a quotient map with the Identity $q\times Id:X\times Z\to Y\times Z$, which is a quotient map if $Z$ is locally compact.
Edit: Actually, showing that $q$ is closed requires only the compactness of $X$. Indeed, if $X$ is compact, then the projection $p_2$ is a closed map, so if $R$ is closed, then for every closed $A\subseteq X$, the set $p_2(A\times X\cap R)$ is closed.
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
Your weaker statement is almost true.
If $f: X \to Y$ is a quotient map and $Z$ is locally compact, then $f \times \operatorname{id}$ is a quotient map. I believe that this result is due to Whitehead.
More generally, if $f: X \to Y$ and $g: Z \to W$ are quotient maps and $Y$ and $Z$ are locally compact, then the product $f \times g: X \times Z \to Y \times W$ is a quotient map.
Why? Use the Whitehead theorem twice, since $f \times g = (\operatorname{id} \times g) \circ (f \times \operatorname{id})$.
See Munkres $\S 22$ for counterexamples.