= Solution
Nonnegativity and symmetry of the <maximum product metric> follow from those of the two factor metrics. Its value is zero precisely when both coordinates agree. For three points $u=(x,x')$, $v=(y,y')$, $w=(z,z')$, the two <triangle inequalities> give
$$
D(u,w)\le\max\{d(x,y)+d(y,z),d'(x',y')+d'(y',z')\}
\le D(u,v)+D(v,w).
$$
All metric axioms therefore hold.
When the two factors are the same <metric space>, consider an off-diagonal point $(x,y)$ and set $\delta=d(x,y)>0$. A product-metric ball of radius $\delta/3$ cannot contain any $(z,z)$: such membership would imply $d(x,z),d(y,z)<\delta/3$ and then $d(x,y)<2\delta/3$, a contradiction. The complement of the diagonal is thus open, so
$$
\boxed{\Delta\text{ is closed in }X\times X.}
$$
This proves the <closed diagonal of a metric space>. Equality of the factors here means equality as metric spaces, as intended; merely sharing an underlying set while allowing unrelated metrics would be a different assertion.
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