Find the interior $\{(x,y): 0 < x^2 + y^2 < 1\}$. 
Find the interior $A = \{(x,y): 0 < x^2 + y^2 < 1\}$. I assume the metric is the standard Euclidean metric.

I know that the int$(A) = A$, but I don't know how to prove it. I can gather that the radius for any $x \in A$ should be $r = \min(\sqrt{x^2_1 + x_2^2},1 -\sqrt{x^2_1 + x_2^2})$ such that $B(x;r) \subseteq A$ by looking at a graph, but how do I show that this is true?

Hint:
It will be easier to write if you think of the plane as the Argand plane.
So let $z_0$ be the affix of the point $M_0(x_0,y_0)$. The affix of a point in the open ball with centre $M_0$ and radius $r$ can be written as
$$z=z_o+u, \enspace \text{where } \;|u|<r.$$
Using the triangle inequality, can you show that, if $\; r<\min(|z_0|,1-|z_0|)$, then $\;0<|z|<1$?
