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I was doing Exercise A.36 in Lee's Introduction to smooth manifolds which states the following:

Let $q: X rightarrow Y$ be an open quotient map. Then $Y$ is Hausdorff if and only if $R = {(x_1, x_2) mid q(x_1) = q(x_2) }$ is closed in $X times X$. I know this has been asked before and my answer agreed with the top answer in this post: $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$. However, I don't see where we use the fact that $q$ is actually a quotient map. It seems that the solution only uses that $q$ is open (so that $q times q$ is open), continuous and surjective. Can someone please clarify?

On a related note, another answer to the same post also states that the product of two open quotient maps is an open quotient map. Is this really true though? I know that in general, the product of two quotient maps is not a quotient map but the counterexamples do not involve maps that are also open.

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 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 we need the information that $q$ is the quotient map.

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.