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.
$\def\RR{\mathbb{R}}$There is no separation condition which will do the job. That's a vague statement, so here is a precise one: There is a subset of $\mathbb{R}^2$ which (equipped with the subspace topology) does not have condition $\dagger$.
Proof: Let $A$ and $B$ be two disjoint dense subsets of $\mathbb{R}$, neither of which contains $0$. (For example, $\mathbb{Q} +\sqrt{2}$ and $\mathbb{Q}+\sqrt{3}$.) Let
$$X = (A \times \RR_{\geq 0}) \cup (B \times \RR_{\leq 0}) \cup (\{0\} \times \RR_{\neq 0}) \subset \RR^2.$$
Define $(x_1,y_1)$ and $(x_2, y_2)$ to be equivalent if $x_1=x_2$ and, in the case that $x_1=x_2=0$, that $y_1$ and $y_2$ have the same sign.
Verification that this is a closed equivalence relation: $X^2$ is a metric space, so we can check closure on sequence. Let suppose we have a sequence $(x_n, y_n) \sim (x'_n, y'_n)$ with $\lim_{n \to \infty} x_n=x$, $\lim_{n \to \infty} y_n=y$, $\lim_{n \to \infty} x'_n=x'$ and $\lim_{n \to \infty} y'_n=y'$. We must verify that $(x,y) \sim (x',y')$. First of all, we have $x_n = x'_n$, so $x=x'$ and, if $x=x' \neq 0$, we are done. If $x=x'=0$, we must verify that $y$ and $y'$ have the same sign. But $y_n$ and $y'_n$ weakly have the same sign for all $n$, so they can't approach limits with different signs.
Verification that $X/{\sim}$ is not Hausdorff: We claim that no pairs of open sets in $X/{\sim}$ separates the images of $(0,1)$ and $(0,-1)$. Suppose such open sets exist, and let $U$ and $V$ be their preimages in $X$. Then there is some $\delta$ such that $(A \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset U$ and $(B \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset B$. Then $U \cap \RR \times \{ 0 \}$ is an open set which contains $(-\delta, \delta)$. By the density of $B$, there must be a point of $B \cap (- \delta, \delta)$ in $U \cap \RR$, and then this gives an intersection between $U$ and $V$.
Best Answer
Wow this is indeed a true statement. I was surprised by how this is not mentioned in basic topology textbooks.
Proof: Suppose $[x],[y]$ are distinct equivalence classes. Since the relation is compact, the classes are also compact and since $X$ is $T_4$, $[x]$ and $[y]$ can be separated by two disjoint open sets, say $U$ and $V$, respectively. Now, we already know that the quotient map $q\colon X \to X/R$ is closed and Proposition 2.4.9 of Enegelking's book on general topology states the following:
For an equivalence relation $E$ over $X$ the following statements are equivalent:
1. The quotient map is closed.
2. For every open set $A\subseteq X$ the union of all equivalence classes that are contained in $A$ is open in $X$.
We can take the union of all equivalence classes contained in $U$ and the union of all equivalence classes contained in $V$ to be $R-$saturated disjoint sets that contain $[x]$ and $[y]$ respectively. This gives that $X/R$ is Hausdorff.