Can open balls of uniform radius around every point of a compact set $S$ be contained in an open set $U$ containing $S$? Let $S$ be a compact subset of a metric space $(X,d)$, and assume $S$ is contained in an open set $U$. Around each point $x$ of $S$, there exists an open ball $B_{\delta'}(x)$ contained in $S$, where the radius $\delta'$ depends on the center $x$.
Question: Define $\delta=\inf_{B_{\delta'}(x)}\delta'$. Is $\delta$ necessarily greater than zero?
Supposing not, for all $\varepsilon>0$ there exists some point $x$ of $S$ such that $B_\varepsilon(x)$ is not a subset of $U$. For each $n\in\mathbb{N}$, take a point $x_n\in S$ such that $B_{1/n}(x_n)$ is not contained in $U$. Then by compactness, the sequence $(x_n)$ has a subsequence $(x_{n_k})$ convergent to a point $x_0\in S$.
Now suppose there exists an open ball $B_\varepsilon(x_0)$ contained in $U$. Take $n_k$ large enough such that $d(x_{n_k},x_0)<\varepsilon/3$, and $x_{n_k}$ satisfies the property that $B_{\varepsilon/3}(x_{n_k})$ is not contained in $U$. Then $d(x_{n_k},x_0)<\varepsilon/3$ implies that $x_{n_k}\in B_{\varepsilon}(x_0)$. But for any point $y$ in $B_{\varepsilon/3}(x_{n_k})$, $d(y,x_0)\leq d(y,x_{n_k})+d(x_{n_k},x_0)=\varepsilon/3+\varepsilon/3=2\varepsilon/3<\varepsilon$, and so $y\in B_{\varepsilon}(x_0)$. This contradicts the assumption that the open ball $B_{\varepsilon/3}(x_{n_k})$ is not contained within $U$, and hence $x_0$ can have no open ball around it contained in $U$. However, this is also a contradiction, since $x_0$ is an element of $S$ and hence an element of $U$, but has no open ball around it contained in $U$.
Is this argument correct? Is the proposition correct to begin with (and if not, does it hold in $\mathbb{R}^n$)? I'm attempting to use this as a step in another proof (that any open cover of the unit circle is also a unit cover of some annulus $\{(x,y)\in\mathbb{R}^2:(1-\rho)^2<x^2+y^2<(1+\rho)^2\}$).
 A: First, a note on notation: $\inf_{B_{\delta'}(x)} \delta'$ doesn't really make sense (although it is clear what you meant to say). You should make explicit the dependence of $\delta'$ on $x$, e.g. by saying "As this is an open cover of $S$, around each $x \in S$ there exists some real number $\delta'(x) > 0$ such that the open ball $B_{\delta'(x)}(x)$ is contained in some element of the subcover". Then you can define $\delta = \inf_{x \in S} \delta'(x)$.
In any case, the answer to your question is (as stated) no. The issue is that the numbers $\delta'(x)$ are not required to "try to be large"! Indeed, for each $x \in S$, there are arbitrarily small positive real numbers $r$ such that $B_r(x)$ is contained in some element of the subcover. This allows us to pick values for $\delta'$ such that $\delta = 0$, as in the following example.
Example. Let $X = \mathbb{R}$ and $S = [0,1]$. Consider the open cover $\{(-1,2)\}$. Choose any bijection $f : S \to (0,1)$. For each $x \in S$, let $\delta'(x) = f(x)-f(x)^2$. Note that $0 < \delta'(x) \leq \frac{1}{4}$ for all $x \in S$, and consequently $B_{\delta'(x)}(x) \subseteq (-1,2)$ for all $x \in S$. However,
$$\delta = \inf_{x \in S} \delta'(x) = \inf_{y \in (0,1)} (y - y^2) = 0.$$
Hopefully this example makes it clear that it's easy to cook up examples where $\delta'$ is chosen poorly, resulting in $\delta = 0$. However, there is a more interesting fact: it is always possible to pick values for $\delta'$ such that $\delta > 0$.
Theorem. Let $(X,d)$ be a metric space and let $S \subseteq X$ be compact. Let $U_1, \dots, U_n$ be open subsets of $X$ such that $\bigcup_{i=1}^n U_i = S$. Then there exists a function $\delta : S \to (0,\infty)$ such that:


*

*For all $x \in S$, there is some $1 \leq i \leq n$ such that $B_{\delta(x)}(x) \subseteq U_i$.

*$\inf_{x \in S} \delta(x) > 0$.


Note: This is a simplified form of the Lebesgue Number Lemma, which answers your strengthened question. The Lebesgue Number Lemma is pretty easy to prove, though, so what follows is a modified proof that applies in this context.
Proof. First, suppose $U_i = X$ for some $i$. Then we may choose $\delta(x) = 1$ for all $x \in S$ and we are done. Otherwise, we have that $Z_i := X \setminus U_i$ is nonempty for all $1 \leq i \leq n$. Let $f : S \to (0,\infty)$ be defined by $f(x) = \frac{1}{n} \sum_{i=1}^n d(x,Z_i)$, where $d(x,Z_i)$ means $\min_{z \in Z_i} d(x,z)$ (note: this is always non-negative). To prove that $f$ is well-defined, let $x \in S$ be arbitrary. There is some $1 \leq i \leq n$ such that $x \in U_i$, so $x \notin Z_i$, whence $d(x,Z_i) > 0$, so $f(x) > 0$. Now, to verify property 2, we simply note that $f$ is continuous (it is the sum of continuous functions), hence (by compactness of $S$) attains a minimum value $D > 0$. Finally, let $\delta : S \to (0,1)$ be defined by $\delta(x) = D$ for all $x \in S$. Since $\delta$ is constant and positive, property 2 is trivially satisfied. To check property 1, let $x \in S$ be arbitrary. Since $f(x) \geq D$, there is some $1 \leq i \leq n$ such that $d(x,Z_i) \geq D$. This means that $D_{\delta(x)}(x) = B_D(x) \subseteq U_i$.
A: A ball contained in the subcover is confounding.
It couldn't mean contained as a subset.
So it must mean an open ball of the subcover  
With that meaning, the construction is impossible for infinite spaces.
The subcover has finite many balls and very few of the balls chosen for each point, will be in the cover.  
Perhaps you intended for each x in S to chose a B(x,r) with radius that is a subset of a ball in the finite subcover.  If indeed you did intend that, there is an example with a zero infinum.   
Let S = [0,1].
C = { B(0,4) } is an open cover and the only finite subcover.
Let K = { 1/n : n in N }.
For all 1/n in K, pick B(1/n,1/2n) and
for all x in S - K, pick B(x,2).  
