If $A$ and $B$ are open in $\mathbb{R}$, show that $A\times B$ is open in $\mathbb{R}^2$. Exercise: Suppose that $A$ and $B$ are open in $\mathbb{R}$. Show that $A\times B$ is open in $\mathbb{R}^2$ with the euclidian metric.
I know that this question pretty much was asked here, but I don't understand the given answer and I would like to know if there's an alternative way.
What I've tried: I know that $A\times B$ is open in $\mathbb{R}^2$, if $\mathbb{R}^2\backslash (A\times B)$ is closed. $\mathbb{R}^2\backslash (A \times B) = (\mathbb{R}\times \mathbb{R}\backslash A)\cup (\mathbb{R}\backslash B \times \mathbb{R})$. I know that $\mathbb{R}\backslash A$ and $\mathbb{R}\backslash B$ are closed, so if I'm able to show that $\mathbb{R}\times \mathbb{R}\backslash A$ is closed, then  I'm done. Unfortunately I don't have a clue on how I should proceed.
Question: How do I solve this exercise?
 A: Take $(a,b)\in A\times B$. Take $r_a>0$ such that $(a-r_a,a+r_a)\subset A$ and take $r_b>0$ such that $(b-r_b,b+r_b)\subset B$. Then $(a-r_a,a+r_a)\times(b-r_b,b+r_b)\subset A\times B$. Let $r=\min\{r_a,r_b\}$. Then the open disk centered at $(a,b)$ with radius $r$ is a subset of $(a-r_a,a+r_a) \times(b-r_b,b+r_b)$ and therefore a subset of $A\times B$. So, $A\times B$ is open.
A: Let $(a_0,b_0) \in A \times B$. Then there is $r>0$ such that 
$|a-a_0|<r$ implies $a \in A$ and $|b-b_0|<r$ implies $b \in B$. Then we have:
$((a-a_0)^2+(b-b_0)^2)^{1/2} < \sqrt{2}r$ implies that $(a,b)  \in A \times B$.
A: I assume the standard topology on $\Bbb R$ and $\Bbb R^2$, so that I can choose any usual metric on $\Bbb R^2$ to prove the exercise. Here I will use $d((x,y),(x',y'))=\max(|x-x'|,|y-y'|)$.
Your edited your question and it says that $\Bbb R^2$ should have the Euclidean metric. However, the maximum metric above and the Euclidean metric generate the same open sets, so my answer still applies. 

Let $(a,b)\in A\times B$. Because $A$ and $B$ are open, there are $\def\eps{\epsilon}\eps_A$ and $\eps_B$ so that $(a-\eps_A,a+\eps_A)\subset A$ and $(b-\eps_B,b+\eps_B)\subset B$. Choose $\eps:=\min(\eps_A,\eps_B)$. Then
$$U_\eps(a,b):=\{(x,y)\in\Bbb R^2\mid d((a,b),(x,y))<\eps\}\subset A\times B.$$
Why is this? Let $(x,y)\in U_\eps(a,b)$. Then
$$|x-a|\le\max(|x-a|,|y-b|)=d((a,b),(x,y))<\eps\le\eps_A\implies x\in(a-\eps_A,a+\eps_A).$$
For the same reason we have $y\in(b-\eps_B,b+\eps_B)$. But then
$$(x,y)\in(a-\eps_A,a+\eps_A)\times(b-\eps_B,b+\eps_B)\subset A\times B.$$
So $U_\eps(a,b)$ is a neighborhood of $(a,b)$ in $A\times B$. And this holds for all points $(a,b)\in A\times B$, hence $A\times B$ is open.
A: Let $(a,b)\in A×B$.
Since $A$ is open: There is a $r$ such 
$B_r(a) \subset A$,
likewise since $B$ is open : There is a $s$ such that
$B_s(b)\subset B$.
Choose $q = \min(s,r)$.
Then 
$B_q(a,b) \subset A×B$ since
for $(x,y) \in B_q(a,b)$:
$((x-a)^2+ (y-b)^2)^{1/2} \lt q$
$\rightarrow :$
$|x-a|\lt q$,  and $|y-b|\lt q$ , i.e.
$x\in A$ and $y \in B$, or
$(x,y) \in A×B$.
