Show that every open cover of the unit sphere $|x|=1$ in $\mathbb R^N$ also covers an annulus $1-\delta\leq|x|\leq1+\delta$ 
Show that, if $S(0,1) \subset \bigcup\limits_{\lambda\in L}A_{\lambda}$ for an open cover $(A_{\lambda})_{\lambda\in L}$, there is $\delta>0$ so that $\bigcup\limits_{\lambda\in L}A_{\lambda}$ covers $K = \{ x \in \mathbb{R}^N \,|\, (1-\delta)^2 \leq |x|^2 \leq (1 + \delta)^2\}$.

So I took $\delta$ in the following way.
$\exists\epsilon>0$, such as $B(a,\epsilon)\subset \cup_{\lambda\in L}A_{\lambda}$, then $\delta = \inf\{\epsilon_a \,|\, B(a,\epsilon_a)\subset \bigcup_{\lambda\in L}A_{\lambda}, a\in S(0,1)\}$
Then I know that I must show $x\in K \Rightarrow x\in \cup_{\lambda\in L}A_{\lambda}$ 
but I'm stuck. Any help?
 A: My hint for what you are struggling with goes as follows: let $d(\cdot)$ be the standard Euclidean metric, i.e. $d(\alpha,\beta)=|\alpha-\beta|$.  Then, by the triangle inequality, $d(x,0)\leq d(x,a)+d(a,0)=d(x,a)+1$, and by the reverse triangle inequality, $d(x,a)\geq|d(a,0)-d(x,a)|=|1-d(x,a)|$.  
But there is a bigger problem with your partial solution: your choice of $\delta$.  How do you know that your $\delta>0$?   You may want to note that $S^{n-1}$ is compact.  
A: It is essential that the unit sphere $S\subset{\mathbb R}^N$ is compact. Define a function
$f:\>S\to{\mathbb R}_{>0}$ by
$$f(x):=\sup\bigl\{\delta>0\bigm|\exists \lambda\in L: \ B_\delta(x)\subset A_\lambda\bigr\}\qquad(x\in S)\ .$$
Claim: This $f$ is continuous.
Proof. Consider a point $x\in S$ and assume $B_\delta(x)\subset A_\lambda$. If $x'$ is a nearby point on $S$ then the triangle inequality implies
$$B_{\delta-|x'-x|}(x')\subset B_\delta(x)\subset A_\lambda\ .$$
Since we can find such an $A_\lambda$ for all $\delta<f(x)$ this implies that $f(x')\geq f(x)-|x'-x|$. This shows that $f$ is even Lipschitz continuous.$\qquad\square$
Since $S$ is compact the function $f$ assumes a positive minimum $\mu$ on $S$. Putting $\delta:={\mu\over2}$ ensures the  claim in the question.
