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 $d_Y$.
(I). We want to show that for any $(p_1,p_2)\in U_1\times U_2$ there exists $r_3>0$ such that the open $d_+$ ball $B_{d_+}((p_1,p_2),r_3)$ is a subset of $U_1\times U_2.$
Take $r_1,r_2>0$ such that $B_{d_1}(p_1,r_1)\subset U_1$ and $B_{d_2}(p_2,r_2)\subset U_2.$ Let $r_3=\min (r_1,r_2).$ Then $$(p,p')\in B_{d_+}((p_1,p_2),r_3)\implies$$ $$\implies (\;d_1(p,p_1)<r_3\leq r_1\;\land\; d_2(p',p_2)<r_3\leq r_2\;) \implies$$ $$\implies (\; p\in B_{d_1}(p_1,r_1)\subset U_1 \;\land \;p'\in B_{d_2}(p_2,r_2)\subset U_2\;)\implies$$ $$\implies (p,p')\in U_1\times U_2.$$
Since this holds for all $(p,p')\in B_{d_+}((p_1,p_2),r_3),$ we have $B_{d_+}((p_1,p_2),r_3)\subset U_1\times U_2. $
(II). Metrics $d$ and $d'$ on a set $X$ produce the same topology iff $$(i).\quad \forall p\in X \;\forall r>0 \;\exists s>0 \;(\;B_d(p,s)\subset B_{d'}(p,r)\;),\text { and }$$ $$ (ii). \quad \forall p\in X \;\forall r>0 \;\exists s'>0\;(\;B_{d'}(p,s')\subset B_d(p,r)\;).$$
To get an idea of how to use this to show that $d_e$ and $d_{max}$ generate the same topology as $d_+,$ consider the case $X_1=X_2=\mathbb R,$ with $d_1(x,y)=d_2(x,y)=|x-y|.$ Sketch some pictures of open balls of various radii, centered at some $p\in \mathbb R^2$, with respect to these 3 metrics.
Metrics on a set $X$ that generate the same topology on $X$ are called equivalent metrics.
Best Answer
Hint: Suppose $X_1 = X_2 = \mathbb{R}$. You know how to make a metric on $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$ out of the metric on $\mathbb{R}$ in a way that looks like the one above (think: Pythagorean theorem). Try and generalize this.