 Solution

The product topology on $X\times Y$ has as a basis the rectangles $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Equivalently, it is the coarsest topology making both coordinate projections continuous.

To show that the product of two Hausdorff spaces is Hausdorff, take distinct points $(x_1,y_1)$ and $(x_2,y_2)$. If $x_1\ne x_2$, choose disjoint open neighborhoods $U_1,U_2$ in $X$; the rectangles $U_1\times Y$ and $U_2\times Y$ separate the product points. If the first coordinates agree, the second coordinates differ, and the same argument uses disjoint open neighborhoods in $Y$. Thus every pair of distinct product points has disjoint open neighborhoods.

The three implications below prove the requested equivalence without assuming metrizability.