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?

This works in any metric space. So you have a distance $d$ and the ball of radius $1$ around $x_0$. 
Let $r=1-d(x_1,x_0)$. If $x$ is at distance less than $r$ from $x_1$, then by the triangle inequality
$$
d(x,x_0)\leq d(x,x_1)+d(x_1,x_0) < r + 1-r = 1. 
$$
In your example  $x_0=(0,0)$, $x_1= (x_1,x_2)$ and $d((a,b),(c,d))=\sqrt{(a-c)^2+(b-d)^2}$.
