Sets of the form $F_1\times\dots\times F_n$, with $F_j$ closed for $1\leq j\leq n$, are closed since their complement is open: it is indeed $$\bigcup_{j=1}^n\left(\prod_{i=1}^{j-1}X_i\times (X_i\setminus F_i)\times \prod_{i=j+1}^nX_i\right).$$
But not all the closed sets have this form: for example, in the plane $\Bbb R^2$, the unit disk $\{(x_1,x_2)\in\Bbb R^2, x_1^2+x_2^2\leq 1\}$ cannot be written as a cartesian product.
(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
$U \subseteq X \times Y$ is open iff for every $(x,y) \in U$ there exist $U_1 = U_1(x) \subseteq X$ open, and $U_2 = U_2(x) \subseteq Y$ with $x \in U_1, y \in U_2$ and $U_1 \times U_2 \subseteq U$.
This just says that all "open squares" are a base for the product topology.
Now check that $\pi_1[U] = \bigcup_{(x,y) \in U} U_1(x)$, which is a union of open sets and thus open.