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.

Draw a picture of the circle that has centre the origin $O$, and radius $1$.
Let $P=(a,b)$ be such that $(OP)^2=a^2+b^2=r^2\gt 1$.  It is obvious that there is an open disk $D$ with centre $P$ such that $D$ lies entirely within $A$.  Now it's over.
But if we want to be explicit, the disk $D$ with radius $\epsilon=\frac{r-1}{2}$ will do the job. For suppose to the contrary that there is an $Q=(x,y)$ such that $x^2+y^2\le 1$ and the distance $PQ$ is less than $\frac{r-1}{2}$.  Then by the Triangle Inequality, we have
$$OP\le OQ +QP.$$
Thus $r\le 1+\frac{r-1}{2}=\frac{1+r}{2}$. This contradicts the fact that $r\gt 1$.
