Is every open cover of $\{(x,y) \in \mathbb R^2 : x^2+y^2=1\}$ also an open cover of some $\delta$ annulus around the unit circle? Let $\{U_\alpha\}$ be an open cover of $\{x \in \mathbb R^n:\|x\|=1 \}$ , $n \ge 2$ , then does there exist $\delta >0$ such that $\{U_\alpha\}$ is also an open cover of $\{ x \in \mathbb R^n : 1-\delta < \|x\| < 1 + \delta \}$ ? 
 A: If that weren't the case you could find $z_n = (x_n,y_n)$ with $1 - {1 \over n} \leq |z_n| \leq 1 + {1 \over n}$ with $z_n \in (\bigcup_\alpha U_\alpha)^c$. Taking a convergent subsequence, you'd have some $z_{n_j}$ converging to a $z$ with $|z| = 1$, but it would also be in $(\bigcup_\alpha U_\alpha)^c$ since the latter set is closed. Hence you have a contradiction.
Another way to look at this: the unit circle is compact, and $(\bigcup_\alpha U_\alpha)^c$ is closed, so there is a positive distance $\delta > 0$ between the two sets, and this $\delta$ can be used.
A: To adapt and complete the OP's initial line of attack, instead of using ordinary unit balls it will make life easier to instead use "polar coordinate basis sets". For each $x = (\cos(\theta_0),\sin(\theta_0)) \in S^1$ there exists $\delta_x > 0$ such that the polar coordinate basis set
$$B_x = \{(r \cos(\theta), r \sin(\theta) \bigm| 1-\delta_x < r < 1+\delta_x, \theta_0 - \delta_x < \theta < \theta_0 + \delta_x\}
$$
is contained in some $U_\alpha$. By compactness of $S^1$, finitely many of the sets $B_{x_1},\ldots,B_{x_K}$ cover $S^1$. Taking $\delta = \text{min}\{\delta_{x_1},\ldots,\delta_{x_K}\}$, the annulus neighborhood 
$$\{(r\cos(\theta),r\sin(\theta) \bigm| 1-\delta<r<1+\delta\}
$$
is contained in the union of the $U_\alpha$'s.
This kind of trick can be seen in the proof of the Tychonoff theorem.
A: At each point $x$ on the sphere, let $\rho(x)$ be the largest number for which the open ball of radius $\rho(x)$ centered at $x$ is a subset of $\bigcup_\alpha U_\alpha$.  The $\rho(x)$ must be finite unless $\bigcup_\alpha U_\alpha=\mathbb R^n$, and if that happens then it's easy to answer the question.
If I'm not mistaken $\rho$ is continuous.  I'd try to write an $\varepsilon$-$\delta$ argument to show this, probably relying heavily on the triangle inequality.
A continuous positive-valued function on a compact space has a minimum value $\delta>0$.
