Prove: if unit sphere is in open set $U$ then $\exists \epsilon>0$ s.t. $\{x\in\mathbb{R^n} | 1-\epsilon \leq d(x,0)\leq 1+\epsilon\} \subset U$ Prove: if the unit sphere $S:=\{x \in \mathbb{R^n} | d(x,0)=1\}$ is in an open set $U \subset \mathbb{R^n}$ (where $\mathbb{R^n}$ is equipped with the Euclidean metric), then $\exists \epsilon>0$ s.t. $C_{\epsilon}:=\{x\in\mathbb{R^n} | 1-\epsilon \leq d(x,0)\leq 1+\epsilon\} \subset U$.
I have tried to solve this using compactness as follows:
$U$, which by definition is a union of open balls: $U = \cup_{\alpha \in I}B(x_\alpha, \epsilon_\alpha)$ is an open cover of $S$. Since $S$ is compact (closed and bounded in $\mathbb{R^n}$) this open cover has a Lesebgue number $\delta >0$. This means that for each $s \in S$, there is some ball $B(x_\alpha, \epsilon_\alpha) \in U$ such that $B(s,\delta) \subset B(x_\alpha, \epsilon_\alpha)$.
My intuition is try and show that setting $\epsilon = \delta$ yields the desired result (at least visually it seems that this could work), but I'm have trouble doing this formally and would appreciated knowing how to do this properly (assuming my approach is indeed correct):
Let $x \in C_\delta$. We can choose $s \in S$ such that $d(x,s) = \min_{s \in S} d(x,s)$.  Then since:
$$d(s,0)-\delta \leq d(x,0) \leq d(s,0)+\delta $$
we obtain:
$$|d(x,0)-d(s,0)|\leq \delta$$
and due to the way we chose $s$ we obtain that [this doesn't feel properly justified]:
$$d(x,s)=|d(x,0)-d(s,0)|$$
and therefore that:
$$d(x,s)\leq \delta$$
meaning $x \in B(s,\delta)$.
Since $x$ was arbitrary, we obtain that for every $x \in C_\delta$ there is an $s \in S$ such that $x \in B(s,\delta)$ and therefore $x \in U$.
Any advice would be much appreciated. Also - if there are other approaches, would be great to know.
 A: It depends how nit-picky you want to be with the definition of Lebesgue number.
Let's go with wikipedia's definition so we're all on the same page.

In topology, Lebesgue's number lemma, named after Henri Lebesgue, is a
useful tool in the study of compact metric spaces. It states:
If the metric space $\ (X,d)\ $ is compact and an open cover of $\ X\ $
is given, then there exists a number $\ \delta >0\ $ such that every
subset of $\ X\ $ having diameter less than $\ \delta\ $ is contained
in some member of the cover. Such a number $ \delta\ $ is called a
Lebesgue number of this cover.

Now you say:

This means that for each $s \in S$, there is some ball $B(x_\alpha, \epsilon_\alpha) \in U$ such that $B(s,\delta) \subset B(x_\alpha, \epsilon_\alpha)$.

The $\ \leq\ $ sign in the definition of $\ C_{\varepsilon}\ $ is annoying here, and we must be more precise.
You should have said:
This means that for each $s \in S$, there is some open ball  $B(x_\alpha, \epsilon_\alpha) \in U$ such that $B(s,\delta) \subset B(x_\alpha, \epsilon_\alpha)\quad (1)$.
If these were closed balls, then $\ C_{\delta}\ $ would do, but because these are open balls, it is not necessarily the case that $\ C_{\delta}\ \subset B(s,\delta).\ $ However, clearly $\ \large{ C_{\frac{1}{2}\delta}\ } $ $ \subset B(s,\delta).$
Now let us continue on from $\ (1).$
Suppose $\ x\in\mathbb{R}^n\ $ is such that $\ 1 - \frac{\delta}{2}\ \leq\ d(x,0)\ \leq 1 + \frac{\delta}{2}.\quad $ Our aim then is to show that $\ \exists s\in S\ $ s.t. $\ x\in B(s,\delta),\ $ i.e. $\ d(x,s)< \delta.\ $ For then we could write
$$ C_{\frac{\delta}{2}} \subset \bigcup\limits_{s\in S} B(s,\delta) \subset \bigcup\limits_{\alpha \in I} B(x_{\alpha},\varepsilon_{\alpha}) = U, $$
and we would be done with the proof.
To this end, suppose $\ x\in\mathbb{R}^n\ $ is such that $\ 1 - \frac{\delta}{2}\ \leq\ \gamma = d(x,0)\ \leq 1 + \frac{\delta}{2}.\quad $ Write
$x = (x_1,\ldots,x_n)\ $ as the coordinates of $\ x\ $ in $\ \mathbb{R}^n.\ $ We have
$$ 1 - \frac{\delta}{2} \leq \sqrt{{x_1}^2 + \ldots + {x_n}^2} \leq 1 + \frac{\delta}{2}.\quad (2) $$
Define $$ s = \left(\frac{1}{\gamma} x_1,\ \frac{1}{\gamma} x_2,\ \ldots,\ \frac{1}{\gamma} x_n \right). $$
$d(s,0)=1,\ $ so $\ s\in S.$ Also, a quick calculation shows that $\ d(x,s) = \vert 1-\gamma \vert \overset{(2)}{\leq} \frac{\delta}{2} < \delta,\ $ and so $\ x\in B(s,\delta),\ $ as desired.
$$$$
Edit: "quick calculation part":
Recall: I use $\ \gamma = d(x,0)\ $ for slight brevity where necessary.
$\ d(x,s) = d\left(\ (x_1,\ldots,x_n),\ \left(\frac{1}{\gamma} x_1,\ \frac{1}{\gamma} x_2,\ \ldots,\ \frac{1}{\gamma} x_n \right) \right)$
$$ = \sqrt{ \left(x_1 - \frac{1}{\gamma} x_1\right)^2  \ldots + \left(x_n - \frac{1}{\gamma} x_n\right)^2 } = \sqrt{ \left( 1 - \frac{1}{\gamma}\right)^2 \left( {x_1}^2 + \ldots + {x_n}^2 \right) } $$
$$ = \left\vert1 - \frac{1}{\gamma}\right\vert \gamma = \left\vert \frac{\gamma - 1}{\gamma}\right\vert \gamma \overset{\gamma>0}{=} \frac{\left\vert\gamma - 1\right\vert}{\gamma}\ \gamma = \left\vert \gamma - 1 \right\vert$$
$
=
\begin{cases}
 \gamma - 1&\text{if}\,\ \gamma > 1\\
 1 - \gamma &\text{if}\,\ \gamma \leq 1\\
\end{cases}
$
$
=
\begin{cases}
 d(x,0) - 1&\text{if}\,\ d(x,0) > 1\\
 1 - d(x,0) &\text{if}\,\ d(x,0) \leq 1\\
\end{cases}
$
Either way, the above together with $\ (2)\ $ implies $\ d(x,s) \leq \frac{\delta}{2} < \delta.$
A: We denote $S:=\{x\in\Bbb R^n|d(x,0)=1\}$ the unit sphere and $C:=C(0;1-r;1+r)=\{x\in \Bbb R^n|1-r\leq d(x,0)\leq 1+r\}$ the cylinder with centre $0$ and radii $1-r, 1+r$. We want to show that $C\subseteq U$.
Firstly, $S\cap\partial U=\varnothing$. If this wasn' t true, there would exist a $y\in S\cap\partial U$; then, provided that $S\subseteq U$, we would obtain that $y\in U$ and $y\in \partial U$, which is false, because $U$ is open, so it contains none of its boundary. Consequently,  $S\cap\partial U=\varnothing$ is true, so $dist(S,\partial U):=δ>0$, which means that $S\subseteq D(0;1+δ)=\{x\in \Bbb R^n|\ ||x||\leq 1+δ\}\subseteq U$. By choosing $\epsilon:=\frac {δ}{2}>0$ we finally obtain that $D(0;1-\epsilon)\subseteq S\subseteq D(0;1+\epsilon)\subseteq D(0;1+δ)\subseteq U$, so $C\subseteq U$, because $D(0;1+\epsilon)\supseteq C=D(0;1+\epsilon)\setminus (D(0;1-\epsilon))^\circ$. q.e.d.
