$A=\{x,y \in \Bbb R $ | $-1 < x < 1, -1< y < 1$ $ \}$ is an open set algebraically. I am trying to show that $A=\{(x,y) \in \Bbb{R} \mid -1 < x < 1, -1< y < 1 \}$ is an open set algebraically. 
Let $a_0 = (x_o,y_o) \in A$. Suppose that $r = \min\{1-|x_o|, 1-|y_o|\}$ then choose $a = (x,y) \in D_r(a_0)$. Then
Edit: I am looking for the proof of the algebraic implication that $\|a-a_0\| = \sqrt {(x-x_o)^2+(y-y_o)^2} < r \Rightarrow|x| < 1 , |y| < 1 $
 A: It is better to choose $r$ smaller than what you mentioned, e.g., $r=\frac{1}{2}\min\{1-|x_o|,1-|y_o|\}$. Well, now starting from $$(x-x_o)^2+(y-y_o)^2<r^2$$since $(y-y_o)^2\geq0$, you can conclude $$(x-x_o)^2<r^2\label{eq1}\tag{1}$$ and since $r<\min\{1-|x_o|,1-|y_o|\}$, you also have $$r^2<(1-|x_o|)^2\label{eq2}\tag{2}$$ By using (\ref{eq1}) and (\ref{eq2}) you'll get $$(x-x_o)^2<(1-|x_o|)^2$$ or equivalently (note that $|x_o|<1$) $$|x-x_o|<|1-|x_o||=1-|x_o|\label{eq3}\tag{3}$$ But using the triangle inequality on the LHS yields $$|x|-|x_o|\leq |x-x_o|$$ Taking this into (\ref{eq3}) gets $$|x|-|x_o|<1-|x_o|$$ or $$|x|<1$$ You can repeat the steps to prove $|y|<1$.
A: Proof by contradiction:
Out of symmetry, Let $ 0 \leq x_o < 1,  0 \leq y_o < 1$
Assumption: $\exists a =(x,y) \in D_r(a_0)$ s.t. $a \not \in A$. Then without loss of generality, let $x>1$( $x<-1$ is impossible, since if $x<-1$, then $|x-x_0| > 1$ which belies the assumption.)
$ 1 - |x_o| \geq r>\sqrt {(x-x_o)^2+(y-y_o)^2} \geq \sqrt {(x-x_o)^2} = |x-x_o|  $
By assumption: $ x-x_o > 1 - x_o \implies |x-x_o| > |1 - x_o|$.
And thanks to the inequality: $|a - b| > |a| - |b|$.
$ |x-x_o| > |1-x_o| > 1 - |x_o|$
Therefore, $1-|x_o| > 1 -|x_o|$ which leads to a contradiction. 
In conclusion, |x|<1 and |y|<1 
