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 the quotient map be $q:X \rightarrow X/A$. $X/A$ is compact because if $\{U_\alpha\}$ is any open cover of $X/A$ then $\{q^{-1}(U_\alpha)\}$ is an open cover of $X$. Since $X$ is compact then there is a finite subcover $\{q^{-1}(U_{\alpha_i})| i = 1,...,n\}$. But then $\{U_{\alpha_i}|i=1,..,n\}$ is a finite subcover of $X/A$ because $q(q^{-1}(U_\alpha)) = U_\alpha$. Therefore, $X/A$ is compact.
To show that $X/A$ is Hausdorff we pick two points $x,y\in X/A$. If $x,y\neq q(A)$ then pick open subsets $U, V \subset X$ such that $q^{-1}(x) \in U$, $q^{-1}(y) \in V, U\cap V = \emptyset, U\cap A = \emptyset, V\cap A = \emptyset$. Then $x\in q(U), y\in q(V)$ and both $q(U)$ and $q(V)$ are open in $X/A$ such that $q(U) \cap q(V) = \emptyset$. If $x \neq y = q(A)$ then you can use the fact that $A$ is compact and Hausdorff to show that there is are open subsets $U, V \subset X$ such that $q^{-1}(x)\in U, q^{-1}(y)\in A\subset V$ and $U\cap V = \emptyset$.
Best Answer
As noted in the comments, this answer shows that $f$ is closed if $R$ is closed in $X\times X$, i.e., that (3) implies (2). Now suppose that $f$ is closed. Note first that since $X$ is Hausdorff, $\{x\}$ is closed for each $x\in X$, and therefore $\{f(x)\}$ is closed for each $x\in X$. And $f$ is a surjection, so $\{y\}$ is closed for each $y\in Y$.
Now let $y_0$ and $y_1$ be distinct points of $Y$, and let $F_i=f^{-1}[\{y_i\}]$ for $i=0,1$; $F_0$ and $F_1$ are disjoint closed sets in $X$. $X$, being compact Hausdorff, is normal, so there are disjoint open sets $V_i$ for $i=0,1$ such that $F_i\subseteq V_i$. For $i=0,1$ let $K_i=X\setminus V_i$, and let $W_i=X\setminus f^{-1}[f[K_i]]$; $f$ is closed and continuous, so $W_i$ is open. It’s easy to check that
$$F_i\subseteq W_i=f^{-1}[f[W_i]]\subseteq V_i$$
for $i=0,1$ and hence that $f[W_0]$ and $f[W_1]$ are disjoint open nbhds of $y_0$ and $y_1$, respectively. Thus, $Y$ is Hausdorff, and (2) implies (1). Since you’ve already shown that (1) implies (2) and (3), the proof that all three are equivalent is complete.