Prove that this set is open $$ A=\left\{ x \in \mathbb{R}^{p} \mid \forall i:\ x_{i} \in (-1,1) \right\}$$

Pick $x\in A$ at random, and choose $\delta = \min B$ where:
$$ B = \{ 1-x_{i}\mid i \in \{1,2,\dotsc,p\} \} \cup \{ 1+x_{i}\mid i \in \{1,2,\dotsc,p\} \} $$
Now pick $y\in \{ z\in \mathbb{R}^p \mid \|x-z\| < \delta \}$.
I have to show that the components $y_i$ are strikly greater than $-1$ and striktly smaller than $1$.
This is where I'm stuck.
The only thing I know is this:
$$ \sqrt{\sum_{i=1}^p(x_i-p_i)^2} < \delta $$
 A: We denote $A_{i}=(-1,1)$ for $1\leq i\leq p$ 
$$
A=\Pi_{i=1}^{p}A_{i}
$$
then if $(x_{i})_{i=1}^{p}\in A$ then $x_{i}\in A_{i}$ but $A_{i}$
is open so there is an open ball $B(x_{i},\epsilon_{i})\subseteq A_{i}$
so by letting $\epsilon:=\min\{\epsilon_{i}\mid1\leq i\leq p\}$ we
get 
$$
B(x_{i},\epsilon)\subseteq B(x_{i},\epsilon_{i})\subseteq A_{i}
$$
so that 
$$
B(x,\epsilon)\subseteq A
$$
so $B$ is an open neighborhood of $x$ contained in $A$, thus $A$
is open.
Note that I have only used the fact that each $A_i$ is open to prove their product is open. 
A: Not really an answer to your question, but a suggestion to follow a different and in my view more elegant route (too much for a comment).
Denote $A_i=\{x\in\mathbb R^p\mid x_i\in (-1,1)\}$ for $i=1,\dots,p$. 
If I understand well then it is your goal to prove that $A=\bigcap_{i=1}^pA_i$ is open in $\mathbb R^p$. For this it is enough to prove that the $A_i$ are open. In that case $A$ is open as finite intersection of open sets. 
Actually the $A_i$ are open by definition when every copy of $\mathbb R$ is equipped with its usual topology and $\mathbb R^p$ is equipped with the producttopology. This because $(-1,1)\subset\mathbb R$ is an open set, and sets of the form $\{x\in\mathbb R^p\mid x_i\in U\}$ with $U$ open in $\mathbb R$ form a subbasis of the producttopology.
A: $$\|x-y\|<\delta \implies \sqrt{\sum_{i=1}^{n}(x_i-y_i)^2}<\delta\implies|x_i-y_i|<\delta\implies|y_i-x_i|<\delta \tag{1}$$
Now, 
$$(\delta\le1-x_i) \land (\delta\le1+x_i)$$
$$\implies (x_i\le1-\delta) \land (-x_i\le1-\delta)$$
$$\implies |x_i|\le 1-\delta \tag{2}$$
Using $(1)$ and $(2)$ and triangle inequality:
$$|y_i|\le|y_i-x_i|+|x_i|<\delta+(1-\delta)=1\ \square$$
