Prove that the hypercube is an open set I'm attempting to prove rigurously that the hypercube is an open set in $\mathbb{R}^p$.
The problem states.

Let $A= \lbrace (x_1,x_2,...,x_p) \in \mathbb{R}^p : 0<x_1<1 , 0<x_2<1 , \dots , 0<x_p<1 \rbrace $. Prove $A$ is open, using the usual metric.  

So far, I was doing this.
I have to prove that $\forall x \in A $ , $\exists r>0$  such that  $B_r(x) \subseteq A$.
That is 
if $x\in \mathbb{R}^p$ then exists $r>0$ such that $B_r (x) \subseteq A$ and if $z\in B_r (x)$ then $z\in A$.
So I proposed $r=1-\max\lbrace x_1,x_2,x_3,...,x_p\rbrace $. That means that $1-\max\lbrace x_1,x_2,x_3,...,x_p\rbrace <x_i $  $\forall i=1,2,...,p$.
Also if $z=(z_1,...,z_p)\in B_r (x)$ then $|z_i-x_i|<d(z,x)<r=1-\max\lbrace x_1,x_2,x_3,...,x_p\rbrace<x_i $. 
So:
$|z_i-x_i|<x_i$ then $-x_i<z_i-x_i<x_i$ for all $i=1,...,p$. Then:
$0<z_i<2x_i<2 $ , $\forall i=1,...,p$. But i wanted $0<z_i<1$ for all $i=1,...,p$ so $z$ would be in $A$. What I've done wrong so far?
Thanks in advance 

Fix $x \in \,]0,1[^p$ and consider the ball $B_r(x)$ with radius
$$
r:=\min\{\min\{x_i,1-x_i\}: i=1,\ldots,p\}.
$$
It follows that, if $z \in B_r(x)$ then, for each $i=1,\ldots,p$, it holds
$$
z_i \in \,\,]x_i-r,x_i+r[ \,\,\subseteq \,\,]0,1[\,.
$$
