Any open cover of $S^1$ is an open cover of the annulas The  question  goes  like  this  : If $\{U_i:i\in I\}$  is  an  open  cover  of  the  unit  circle  in  $\mathbb R^2$ then  show  that  it  is  an  open   cover  of  an  annulus $1-\delta\lt ||(x,y)||\lt 1+\delta$  for  some  $\delta \gt 0$.
 I have  two  problems  with  this  statement :
$1)$ I can understand the statement  with   finite  number of  covers geometrically  but  with  infinite  number  of  covers , I  find  it  difficult  to  believe . 
$2)$ Also  whatever  I  understand  is  completely  geometric and  not  analytic. 
All  the  open  sets  that  cover  the  circle $S^1$  are  open  discs  in  $\Bbb R^2$ .  Since  $S^1$  is  compact , we  have finite  number  of  them  covering  the  whole  of  $S^1$ .  Say  they  are  $D_1,D_2,.....D_k.$  And  without  loss  of  generality ,  let  $D_m$  be  the  one with  the  smallest  radius,$\mathcal r$. { For convenience, can  we  consider  only  circles  that  have  their  centers  on  $S^1$ $?$ }  Then  we can easily  see  that  $\{D_i;i=1,2,...k\}$  cover  the  annular  region $D(0,1+\mathcal r)\backslash D(0,1-\mathcal r)$
[Here  I  considered  discs  only  for  I  thought  that any  open  set  has  an  open  disc  embedded  in  it  so  we  could  take  into  account  the  sub-cover  consisting  of  the  open  sets  that  contain  ourchosen  open  discs ]
So if  I  have  to  consider  all   the  infinite   number  of discs  covering $S^1$ , then  there  is  a  problem  for  it  can  be  the  set  of  circles  of  radius  $1\over n$ . So  the  infimum  is  $0$  although  none  has  radius  $0.$ So  finding  the  one  with  smallest  radius  is  impossible  here  so  my  trick  won't  work  here. So  how  do  we  get  the  result  for  infinite  case $?$
And  from  the  look  of  the  question  it  seems  considering  the  whole  cover  and  not  a  finite  sub cover  is  desired  here. 
So please  help  me  with  the  proof .
Thanks.
 A: First I'll consider your (bolded) question about taking only discs with centers in $S^1$. Indeed we can do this - consider your open cover $\{ U_i\}$. Each $x\in S^1$ is contained in some $U_x\in\{ U_i\}$, and $U_x$ is open, so it contains some open ball in $\mathbb{R}^2$ centered at $x$, say $B_x$. Indeed, it should be clear that the collection of open balls $\{ B_x\}_{x\in S^1}$ form an open cover of $S^1$ (since given $x\in S^1$, $x\in B_x$). Therefore, since $S^1$ is compact, we can find a finite subcover $\{ B_1,\ldots B_n\}$, which are open balls in $\mathbb{R}^2$ with centers in $S^1$ as you desired.
However, your comment that the minimal radius works is, in fact, incorrect. Indeed, by considering discs that only just overlap one another you should be able to draw a fairly convincing picture to illustrate this (the problem lies in the "hollows" between the discs, which may approach closer to the circle than the minimum radius of any disk). Instead, you need to consider the distance of $S^1$ from the complement of the union of sets in our cover. 
Let $C:=(\bigcup\limits_{j=1}^n B_j)^\mathsf{c}$. This is a closed set, and therefore we can find a $\delta>0$ such that $$d(S^1,C):=\inf\{\|x-c\|\mid x\in S^1, c\in C\}>\delta.$$ Indeed, if we could not then we could find a sequence in $C$ converging to an $x\in S^1$, which would tell us that $x\in C$ (since $C$ is closed). This is a contradiction because the complement of $C$ was meant to cover $S^1$. 
Now this $\delta$ works.
