Prove that $S= \{ (x,y) : x^2 - y^2 <1 \}$ is open in $\mathbb{R}^2$ 
Prove that $S= \{ (x,y) : x^2 - y^2 < 1 \}$ is open in
  $\mathbb{R}^2.$

The question itself is rather easy and trivial by observing the $S$ in $\mathbb{R}^2$ geometrically. But if we are restricted to prove the above by using the definition of open set ONLY, how should we write down the proof formally?
A set $P$ is called open if for every $x \in P$, there exists $r>0$ such that $B(x, r) \subset P$.
Particularly, in this question, what value of $r$ should we choose for a given $(x,y) \in S$?
Thanks in advance.

Consider the function $f(x,y) = x^2 - y^2$, $f\colon \mathbb{R}^2 \to \mathbb{R}$. It is certainly continuous. Then $S = f^{-1}( (-\infty, 1) )$. Since $(-\infty, 1)$ is open, $S$ is open too.
As for your question about proving openness from the very definition, note that the boundary of $S$ is just a hyperbola. For any point in $S$ you have enough room to find a ball around that point between the branches of this hyperbola.
