= Solution
The <product topology> on $X\times Y$ consists of arbitrary unions of rectangles $U\times V$ with $U\subseteq X$, $V\subseteq Y$ open. Such rectangles form a <basis of a topology>: every point lies in one, and intersections of two rectangles are rectangles with open factors.
If $d_X,d_Y$ induce the factor topologies, set
$$
\boxed{d((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')\}.}
$$
Definiteness and symmetry are immediate. Each factor <triangle inequality> bounds its distance by the sum of the two relevant product distances; taking the maximum proves the product <triangle inequality>. A ball of radius $r$ is exactly $B_X(x,r)\times B_Y(y,r)$, so every metric-open set is product-open. Conversely an open rectangle containing $(x,y)$ contains a product ball with radius the minimum of two suitable factor radii. Hence the <maximum product metric> induces precisely the <product topology>.
A <compact space> is one for which every <open cover> has a finite subcover. Let $\mathcal U$ cover $X\times Y$, with both factors compact; if either is empty the result is immediate. For each fixed $x$ and each $y$, choose a rectangle $U_{x,y}\times V_{x,y}$ containing $(x,y)$ and lying in a member of $\mathcal U$. The sets $V_{x,y}$ cover $Y$, so finitely many, indexed by $y_1,\ldots,y_{m_x}$, suffice. Their $X$-neighbourhoods have open intersection
$$
U_x=\bigcap_{j=1}^{m_x}U_{x,y_j},
$$
containing $x$. The finitely many selected members of $\mathcal U$ cover all of $U_x\times Y$. Now the sets $U_x$ cover $X$, so select finitely many of these by compactness of $X$. Taking the union of their finite selections from $\mathcal U$ gives a finite cover of the whole product. \b[The product of two compact spaces is compact], without requiring the factors to be Hausdorff.
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