 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: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 closedness of the relation and openness of the quotient map suffice for a Hausdorff quotient.