As for (2), it suffices to prove that the balls produced by the distance $d_{X\times Y}$ are a basis for the product topology.
So let's write down what it is a ball for the distance $d_{X\times Y}$ with center $(a,b)$ and radius $\varepsilon > 0$:
$$
B_{X\times Y}((a,b); \varepsilon) = \left\{ (x,y) \in X\times Y \ \vert \ d_{X\times Y} ((a,b), (x,y))= \max \left\{ d_X(a,x) , d_Y (b,y) \right\} < \varepsilon \right\} \ .
$$
But
$$
\max \left\{ d_X(a,x) , d_Y (b,y) \right\} < \varepsilon \quad \Longleftrightarrow \quad d_X(a,x) <\varepsilon \ \text{and} \ d_Y(b,y) < \varepsilon \ .
$$
Hence we see that
$$
B_{X\times Y }((a,b); \varepsilon)\ = \ B_X (a;\varepsilon ) \times B_Y (b; \varepsilon ) \ .
$$
That is: a ball for the distance $d_{X \times Y}$ is the same as a product of balls for the distances $d_X$ and $d_Y$. Which means that balls for the distance $d_{X\times Y}$ form a basis for the product topology.
You have to prove:
$$\begin{eqnarray*}d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2 &\leq& d_X(x_1,x_2)^2+d_X(x_2,x_3)^2+d_Y(y_1,y_2)^2+d_Y(y_2,y_3)^2\\&+&2\sqrt{\left(d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2\right)\left(d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2\right)}\end{eqnarray*}$$
where the Cauchy-Schwarz inequality ensures that the last square root is greater or equal than:
$$ d_X(x_1,x_2)\,d_X(x_2,x_3)+d_Y(y_1,y_2)\,d_Y(y_2,y_3)$$
hence it is sufficient to show that:
$$d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2\leq\left(d_X(x_1,x_2)+d_X(x_2,x_3)\right)^2+\left(d_Y(y_1,y_2)+d_Y(y_2,y_3)\right)^2$$
that just follows from the triangle inequality for $d_X$ and $d_Y$.
Best Answer
You can define, for instance,$$d\bigl((x_1,y_1),(x_2,y_2)\bigr)=\max\bigl\{d_X(x_1,x_2),d_Y(y_1,y_2)\bigr\}$$or$$d\bigl((x_1,y_1),(x_2,y_2)\bigr)=d_X(x_1,x_2)+d_Y(y_1,y_2),$$or even$$d\bigl((x_1,y_1),(x_2,y_2)\bigr)=\sqrt{d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2}.$$Any of them will do. And they all induce the product topology.
But it is not true that $X\times Y$ is complete with respect to any distance that you define on it, even if it induces the product topology.