Prove that $(a,b)\times(c,d)\subset \mathbb{R}^2$ is an open set. Let $\|\cdot\|$ be a norm of $\mathbb{R}^2$.
Prove that
$$\forall x\in (a,b)\times(c,d)\subset\mathbb{R}^2,\; \exists \varepsilon&gt;0,\; B(x,\varepsilon)\subseteq (a,b)\times (c,d)$$
where $\displaystyle B(x,\varepsilon)=\{v\in \mathbb{R}^2: \|v - x\|&lt;\varepsilon \}$.

My attempt:
Let $x=(x_1,x_2)\in (a,b)\times (c,d)$. So $a&lt;x_1&lt;b\land c&lt;x_2&lt;d$.

Figure suggests that the radius of the open ball (blue) ought be smaller than any of the green lines (or equal to the shortest one).
Let $\displaystyle\varepsilon =\min\{\|(x_1,d)-x\|,\|(x_1,c)-x\|,\|(a,x_2)-x\|,\|(b,x_2)-x\|\}=\|\alpha-x\|$.
And let $v\in B(x,\varepsilon)$, that is
$$\|v-x\|&lt;\|\alpha-x\|.\tag{*}$$
I can't find a way to prove that $(*)$ implies $v\in (a,b)\times (c,d)$.
How could I continue the proof? If I'm not going in the right direction a hint would also help.
On a finite product,  $\mathbb R\times\mathbb R $, in this case,  the product and box topology coincide.  So the cross product of two open sets is open, by definition.
Put differently,  if $\pi_1,\pi_2$ are the projections onto the first and second factor, then $(a,b)×(c,d)=\pi_1^{-1}(a,b)\cap\pi_2^{-1}(c,d) $.
On a finite product,  $\mathbb R\times\mathbb R $, in this case,  the product and box topology coincide.  So the cross product of two open sets is open, by definition.
Put differently,  if $\pi_1,\pi_2$ are the projections onto the first and second factor, then $(a,b)×(c,d)=\pi_1^{-1}(a,b)\cap\pi_2^{-1}(c,d) $.