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.
First note (or know, or prove) that in $\mathsf{Top}$ a monomorphism is a continuous injection (and an epimorphism is a continuous surjection), see Wikipedia e.g.
So write $q = m \circ f$ where $m:Y \to Z$ and $f: X \to Y$ is continuous and $q: X \to Z$ being the given quotient map (or take $Z=X{/}{\sim}$ if you like) and $m$ is an injective continuous map. We need to show that $m$ is a homeomorphism (i.e. an isomorphism in $\mathsf{Top}$).
So $m$ needs to be open and surjective as well. Openness is equivalent to the inverse being continuous (for bijections).
$m$ is surjective: if $z \in Z$ we find $x \in X$ with $q(x)=z$ (quotients maps are surjective) and then $m(f(x)) = q(x)= z$, showing that $m$ is surjective.
Openness of $m$: let $O$ be open in $Y$ and note that $m[O]$ is open in $Z$ iff $q^{-1}[m[O]]$ is open in $X$ (by definition of the quotient topology/quotient maps) and simple set theory tells us:
$$q^{-1}[m[O]] = (m \circ f)^{-1}[[m[O]] = f^{-1}[m^{-1}[m[O]] = f^{-1}[O]$$
where the last step uses injectivity of $m$. So $q^{-1}[m[O]]$ is open by continuity of $f$ and so $m[O]$ is open and we are done.
Best Answer
Your map $(q,g)$ is a dual to the diagonal map into products (if we have maps $f_i : X \to Y_i$ with a common domain, $\nabla_i f_i: X \to \prod_i Y_i$ defined by $\pi_i \circ \nabla_i f_i = f_i$ for all $i$ is also continuous), where here we have a common codomain $Y$ and we define $(q,g)$ by $(q,g) \circ j_X = q$ and $(q,g) \circ j_A = g$ and the universal property for sums implies that $(q,g)$ is continuous (it has the final topology wrt the embeddings $j_X: X \to X \oplus A$ and $j_A: A \to X \oplus A$).
That $(q,g)$ is quotient can also be seen by the definitions: indeed it is surjective as $q$ is (just as $\nabla_i f_i$ is injective whenever just one $f_i$ is) and if $(q,g)^{-1}[A] = q^{-1}[A] \oplus g^{-1}[A]$ is open in $X \oplus A$ this means that $q^{-1}[A]$ is open in $X$ and $g^{-1}[A]$ is open in $A$. The first part already implies that $A$ is open in $Y$ as $q$ is quotient, and we are done.