Solution: We will show that $X$ is Hausdorff and replacing the arguments for $Y$ will show that $Y$ is also Hausdorff. Let $x_1, x_2 $ be two distinct points from $X$. Take $y_0 \in Y$. Consider the points $\left( x_1, y_0 \right) $ and $\left( x_2, y_0 \right) $. These points are distinct and $X \times Y$ is hausdorff, so we can find disjoint open sets $U_1 \times V_1$ and $U_2 \times V_2$ confining points $\left( x_1, y_0 \right) $ and $\left( x_2, y_0 \right) $ respectively. We claim that $U_1 \cap U_2 = \emptyset $. If not, let $x \in U_1 \cap U_2$. Then, $(x,y_0) \in U_1 \times V_1 \cap U_2 \times V_2$, a contradiction. Thus, $X$ is Hausdorff.