Proving that every bounded closed and open subset of an affine space over an locally compact ultrametric field is a disjoint union of balls Let $k$ be a field complete with respect to a nonarchimedean absolute value $|\cdot|$, such that the resulting topology on $k$ is locally compact.
Let $x\in k^n$, and let $B_r(x)$ denote the ball of radius $r$ ($r > 0\in\mathbb{R}$) in $k^n$.
Serre's lecture notes on Lie Algebras and Lie Groups has a Lemma (p98):
Lemma: Let $U$ be a closed and open set of a ball $B$ in $k^n$. Then there is a positive radius $r$ smaller than the radius of $B$ such that $U$ is the disjoint union of a finite number of balls of radius $r$.
His proof: Let $V := B - U$. Then $\{U,V\}$ is an open covering of the compact metric space $B$. Hence there is a radius $r$ less than the radius of $B$ such that, for all $x\in B$, the ball of radius $r$ about $x$ in $B$ is contained in either $U$ or $V$. By the preceding remark (where he proves that two balls in $k^n$ are either contained one in the other, or are disjoint), we see that a ball of radius $r$ in $B$ is a ball of radius $r$ in $k^n$. Hence $U$ is the union of balls of radius $r$ in $k^n$.
How does one deduce his second sentence? (the existence of the $r$)?
(Earlier he also proves that all balls are open and compact).
 A: For each $x \in B$, there is an $r_x > 0$ such that the ball of radius $r_x$ about $x$ is contained in $U$ (if $x \in U$) or in $V$ (if $x \in V$), since $U$ and $V$ are open. These balls, for $x \in B$, cover $B$. Since $B$ is compact there is a finite subcover, say with centers $x_1,\dotsc,x_s$. Let $r$ be the smallest radius occuring in this subcover.
We claim that for every $x \in B$, the ball $B_r(x)$ is contained in $U$ (if $x \in U$) or in $V$ (if $x \in V$). Say $x \in U$. There is an $i$, without loss of generality $i=1$, so that $x \in B_{r_1}(x_1)$, and $r_1 \geq r$. And also $B_{r_1}(x_1) \subseteq U$.
But now $B_r(x) \subseteq B_{r_1}(x_1)$, since the metric on $k^n$ is an ultrametric and $r \leq r_1$. So $B_r(x) \subseteq U$.
(This is far more basic than your question, but just to be sure: if $y \in B_r(x)$, then $\|y-x\| < r$. And $\|x-x_1\| < r_1 \leq r$. So
$$
  \|y-x_1\| = \|(y-x)+(x-x_1)\| \leq \max\{\|y-x\|,\|x-x_1\|\} < r
$$
which shows $y \in B_{r_1}(x_1)$.) (I hope you don't mind. I wrote that as much for myself as for you, since I haven't thought about such things in some time.)
