Proving $(A\times B)^- = A^-\times B^-$ (closure of cartesian product) My proof, for:
$$(A\times B)^- = A^-\times B^-$$
using the metric 
$$d''((a_1,a_2),(b_1,b_2)) = max\{d_1(a_1,b_1),d_2(a_2,b_2)\}$$
$\rightarrow$
Well, if $a = (a_1,a_2)\in (A\times B)^-$ then:
$$d''((a_1,a_2), A\times B) = 0 \implies d''((a_1,a_2),(a_a,a_b))=0 \implies d_1(a_1,a_a)=0, d_2(a_2,a_b)=0$$
for all $a_a\in A$ and $a_b\in B\implies a_1\in A^-$ and $a_2\in B^-$, therefore $a\in (A\times B)^-$
$\leftarrow$
$$a\in A^- \times B^ \implies (a_1,a_2)\in A^- \times B^- \implies a_1\in A^-, a_2\in B^-\implies d_1(a_1,a_a)=0 \ \  \forall a_a\in A, \\d_2(a_2,a_b)=0 \ \ \forall a_b\in B \implies d''((a_1,a_2),(a_a,a_b))=0, \ \ \forall a_a\in A, a_b\in B\implies d''(a,A\times B)=0\implies a\in (A\times B)^-$$
 A: The distance of a point to a set is defined as $d(a,B) = \inf\{d(a,b)\mid b\in B\}$. So you cannot conclude that from $d(a,B) = 0$ we have $d(a,b) = 0$ for all $b\in B$. All you can say is that for every $\varepsilon>0$ there exists $b\in B$ such that $d(a,b)<\varepsilon$.
Use the property: $a\in \overline A$ if and only if for all $\varepsilon>0$ there exists $b\in A$ such that $d(a,b)<\varepsilon$.
So your proof should go like this: Let $(a,b)\in \overline{A\times B}$. We need to show $a\in \overline A$ and $b\in \overline B$. So let $\varepsilon>0$ be arbitrary. Then there exists $(a',b')\in A\times B$ such that $d''((a,b),(a',b')) = \max\{d_1(a,a'), d_2(b,b')\} < \varepsilon$. Hence $d_1(a,a')<\varepsilon$ and $d_2(b,b')<\varepsilon$. Since $\varepsilon>0$ was arbitrary, we conclude $a\in \overline A$ and $b\in \overline B$, i. e. $(a,b)\in \overline A\times \overline B$.
Conversely, let $(a,b)\in \overline A\times \overline B$. Let $\varepsilon>0$ be arbitrary. Since $a\in \overline A$, there exists $a'\in A$ such that $d_1(a,a')<\varepsilon$. Since $b\in \overline B$, there exists $b'\in B$ such that $d_2(b,b')<\varepsilon$. But then we have
$$
d''((a,b),(a',b')) = \max\{d_1(a,a'), d_2(b,b')\} < \varepsilon.
$$
Since $\varepsilon>0$ was arbitrary, we conclude $(a,b)\in \overline{A\times B}$.

Notice, that the inclusion $\overline{A\times B}\subseteq \overline A\times \overline B$ is actually almost trivial, because $\overline A\times \overline B$ is closed (its complement being $((X\setminus A)\times Y)\cup (X\times (Y\setminus B))$, a union of open sets; here $X\supseteq A$ and $Y\supseteq B$ are the total spaces) and $\overline {A\times B}$ is the smallest closed set containing $A\times B$ (i. e. the intersection of all closed sets containing $A\times B$).
