Proving that a set is open using epsilons. I'm trying to prove that the set $$A=\{x=(x_{1},x_{2})\in\mathbb{R}^2:x_{1}^{2}+x_{2}^{2}>1\}$$ is open in $\mathbb{R}^2$ with the usual norm is open with the definition of "epsilons". 
My attempt is based in taking $\epsilon=x_{1}^{2}+x_{2}^{2}-1,$ but simply don't achieve to prove that $B(x,\epsilon)\subset A.$ I tried too using another norm because of the equivalence among norms, but it's useless. 
Another form to attack the problem is taking complement over A, so $A^c$ is closed that's why is an closed ball. However I'd know the path using epsilon definition.
Anyone could help me?
Thanks in advance.

Let $\epsilon = d(x, 0) - 1$. Thus $d(x, 0) - \epsilon = 1$ where $d$ is the standard metric. 
Then if $y \in B_\epsilon(x)$ (open ball of radius $\epsilon$ centered at $x$), we want to show $d(y, 0) > 1$ (since $d(y, 0) > 1$ if and only if $y_1^2 + y_2^2 > 1$). 
By the triangle inequality 
$$d(y, 0) + \epsilon > d(y, 0) + d(x, y) \ge d(x, 0) > 1$$
Thus $$d(y, 0) > d(x, 0) - \epsilon = 1.$$
