Show that an "open square" is an open set: Show that $\{(x,y) \in \mathbb{R}^2$ such that $-1How do I prove that an "open" square, centered in the origin is in fact an open set? I've already have this geometrical argument:
Let $S$ denote the square.
Suppose $(x,y) \in S$. Let $\delta = \min \{1 - |x|, 1 - |y|\}$. 
Then, geometrically it is clear that $B_\delta(x,y) \subseteq S$. Hence $S$ is open. 
However, How can I write this down and prove it in a formal matter?

Let $S$ denote the square you wrote in the comments. 
Suppose $(x,y) \in S$. Let $\delta = \min \{1 - |x|, 1 - |y|\}$. 
Then $B_\delta(x,y) \subseteq S$. Hence $S$ is open. 
