Suppose that the family $ \mathcal{F}$ is a family of open sets in $ \mathbb{R}^2$ which covers the circle $C=\{(x,y): \ x^2+y^2 = 1 \}$. Show that there is a $ \rho>0$ such that $ \mathcal{F}$ covers the set
$A=\{(x,y): \ (1-\rho)^2 \leq x^2+y^2 \leq (1+\rho)^2 \}$.
Answer:
If we show that the family $ \mathcal{F}$ covers the set $ \{(x,y): 0 \leq x^2+y^2 \leq (1+\rho)^2 \}$, then it will be sufficient.
Since the family $ \mathcal{F}$ covers the unit circle, the covers contains sets of radius $ 1+ \rho \ $ for some $\rho>0$ (without loss of generality).
Thus $ \mathcal{F}$ covers the set $A$.
But I think, it is not appropriate .
Help me
There is no reason why points close to the origin are covered, so your argument is not correct. Let us prove this by contradiction. Suppose for any $\rho$ there exists poits in the annulus $A$ which are not covered. Then there is a sequence $(x_n,y_n)$ with $x_n^{2}+y_n^{2} \to 1$ such that no $(x_n,y_n)$ belongs to any of the open sets in our open cover. By Bolzano - Weirstrass Theorem there is a subsequence which converges to some point $(x,y)$. Since $x^{2}+y^{2}=1$ the point $(x,y)$ belongs to some open set $U$ in the open cover. But then $(x_n,y_n)$ is also in $U$ for $n$ sufficiently large. This is a contradiction.