Let $S^1=\{x\in\mathbb{R}^2\mid\lVert x\rVert=1\}$, where $\lVert x\rVert$ denotes the Euclidean norm. I am asked tho prove that if $U$ is an open set, $S^1\subset U$, then there exists $\epsilon>0$ such that the set $\{x\in\mathbb{R}^2\mid \lVert x\rVert>1-\epsilon,\lVert x\rVert<1+\epsilon\}$ is contained in $U$.
I tried using the compactness of $S^1$.
$\forall x\in S^1,\;\exists\,\delta_x>0 \mid B(x,\delta_x)\subset U$. So the set $\{B(x,\delta_x)\mid x\in S^1\}$ is an open cover of $S^1$. $S^1$ compact implies there exists a finite subcover, for example, $B(x_1,\delta_{x_1}),\ldots,B(x_k,\delta_{x_k})$.
Is it enough to take $\epsilon=\min\{\delta_{x_1},\ldots,\delta_{x_k}\}>0$?
I'll expand a little bit on my hint.
Given two closed sets $A$ and $B$, one of which is bounded, define $$ d(A,B) = \inf \{ d(a,b) : a \in A, b \in B \} $$ Then $d(A,B) \geq 0$ and $d(A,B) = 0$ if and only if $A \cap B \neq \varnothing$.
Hence, if you can prove that $\partial U \cap S^1 = \varnothing$, then you are done...