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?

The interval $(2,4)$ is open in $\Bbb R$, and the function $f(x,y)=x^2+y^2$ is continuous, thus as the preimage of an open set under a continuous function, the said set is continuous.
