= Solution
A <Hausdorff space> separates any two distinct points by disjoint <open neighborhoods>. For a subspace $A\subseteq X$, intersect such neighborhoods in $X$ with $A$ to obtain disjoint neighborhoods in the <subspace topology>. Thus subspaces inherit the <Hausdorff> property.
If $(z_1,z_2)$ lies off the diagonal of a <Hausdorff space>'s square, choose disjoint <open neighborhoods> $U$ of $z_1$ and $V$ of $z_2$. The <open> rectangle $U\times V$ contains the pair and avoids the diagonal. Therefore its complement is <open> and the diagonal is <closed>.
The map $F:X\times Y\to Z\times Z$ given by $F(x,y)=(f(x),g(y))$ is <continuous> by the coordinate criterion from E3. The equality locus is $F^{-1}(\Delta_Z)$ and is <closed> as an <inverse image> of a <closed set>.
Let $q:X\to X/{\sim}$ be the <quotient map>. If the <quotient> is <Hausdorff>, apply the same argument to $q$ in both coordinates:
$$
R=(q\times q)^{-1}(\Delta_{X/{\sim}})
$$
is <closed> in $X\times X$.
Conversely suppose $R$ is <closed> and $q$ is <open>. Take two distinct <quotient> points, with representatives $x,y$. Since $(x,y)\notin R$, an <open> rectangle $U\times V$ around it avoids $R$. The <images> $q(U)$ and $q(V)$ are <open> <quotient> neighborhoods. They are disjoint, because a <quotient> point in their <intersection> would have equivalent representatives $u\in U$, $v\in V$, giving $(u,v)\in R$. Hence the <quotient> is <Hausdorff>. Thus \b[closedness of the relation and openness of the <quotient map> suffice for a <Hausdorff> <quotient>].
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