Prove that the set $\{(x,y) \in\mathbb R^2\mid 2I've been trying to prove this in the following way:
Name the area between the two circles $S$
Suppose $(x,y) \in S$. Let $\delta = \min \{\sqrt{4-x^2-y^2},\sqrt{x^2+y^2-2}\}$. 
Then, I'm trying to show that $B_\delta(x,y) \subseteq S$, i.e: for $(u,v)$ in $B_\delta(x,y)$ it is true that $2<u^2+v^2<4$. 
Is this approach right? How can I write this down and prove it in a formal matter?

You can prove this more easily:
If $A$ and $B$ are opens, then $A\cap B$ is open.
note that 
$$S=B(0,4)\cap(\Bbb{R}^2- B[0,2]),$$
where $B(0,4)$ is the open ball in $(0,0)$ of radius $4$, and $B[0,2]$ the closed ball in $(0,0)$ with radius $2$.  
