If a compact set is covered by a finite union of open balls of same radii, can we always get a lesser radius?

This question seems obvious, but I'm not secure of my proof.

If a compact set $V\subset \mathbb{R^n}$ is covered by a finite union of open balls of common radii $C(r):=\bigcup_{i=1}^m B(c_i,r)$, then is it true that there exists $0<s<r$ such that $V\subseteq C(s)$ as well? The centers are fixed.

I believe this statement is true and this is my attempt to prove it:

Each point of $v\in V$ is an interior point of least one ball (suppose its index is $j_v$), that is, there exists $\varepsilon_v>0$ such that $B(v,\varepsilon_v)\subseteq B(c_{j_v},r)$, so $v\in B(c_{j_v},r-\varepsilon_v)$. Lets consider only the greatest $\varepsilon_v$ such that this holds. Then defining $\varepsilon:=\inf\{\varepsilon_v\mid v\in V\}$ and $s=r-\varepsilon$ we get $V\subseteq C(s)$.

But why is $\varepsilon$ not zero? I thought that considering the greatest $\varepsilon_v$ was important, but still couldn't convince myself.

I would appreciate any help.

Replace each open ball $B_i$ of radius $r$ in the cover by the union of concentric open balls of radii strictly smaller than $r$. You get an infinite cover of $V$. By compactness there is a finite subcover. By construction the radii are smaller than before. Finally we choose the maximal radius (for all of the finitely many balls) which is still smaller than $r$.

Let $X$ denote the set of centers: $X = \{c_1,\ldots,c_m\}$.

The function $\phi(x) = \mathop{\rm dist} (x,X)$ is continuous on $\mathbb R^n$ and attains a maximum value on $V$ because $V$ is compact.

Note that if $x \in V$, then by definition $\phi(x) < r$. Whatever maximum it attains must be less than $r$.

Choose $s$ to lie in between this maximum and $r$.