Ultrametric ball I am trying to prove that any ball in ultrametric space is the union of some finite number of smaller balls. I can't find proof and don't unerstand how to do it. I thought to bring the explicit form of splitting into 2 balls, but I did not succeed. I understand that ultrametric balls don't intersect. We can introduce an equivalence relation and our ball will be a union of disjoint balls.
$$
\mathrm{B} \left(a, \, R \right) = \bigcup_{x \, \in \, \mathrm{B} \, \left(a, \, R \right)} \mathrm{B} \left(x, \, \varepsilon \right)
$$
But how to prove that their number is finite? Thank you!

Added. (for ultrametric spaces, in which all closed ball is compact):
We know that all balls in the ultrametric space are clopen. Then we can introduce a binary relation so that $ x \sim_{r} y $ if $y \in \mathrm{B} \left(x, \, r \right) $, where $\mathrm{B} \left(a, \, r \right) = \left\lbrace x \in \mathrm{X}| \, \mathrm{d}(x, a) \leq r  \right\rbrace $. This is a binary relation, because the properties of reflexivity, symmetry, and transitivity are satisfied. Since every equivalence relation generates a partition of the space into disjoint equivalence classes, then
$$
\mathrm{X} = \bigcup_{x \, \in \, \mathrm{X}} \mathrm{B} \left(x, \, r \right)
$$
for any $r$. So any ball will be a union of disjoint balls, i.e
$$
\mathrm{B} \left(a, \, R \right) = \bigcup_{x \, \in \, \mathrm{B} \, \left(a, \, R \right)} \mathrm{B} \left(x, \, r \right).
$$
If any ball is compact, then in each of its open covers has a finite subcover. In this space balls are clopen, then in our cover of balls has finite subcover. Thus any ball is the union of a finite number of smaller balls. Is it correctly?
 A: It is not true. To see this, consider the metric on the positive integers
where $d(m, n) = d(n, m) = 1 - 2^{-m}$ when $m > n$. This is an ultrametric
because we have $d(m, n) = d(m, p) > d(n, p)$ whenever $m > n > p$.
The open unit ball about any point is the whole space, but for any radius
$0 < r < 1$ we have $B(n, r) = \{n\}$ for all $n > -\log_2 (1-r)$.
It follows that every ball smaller than a unit ball is a finite set, so no 
finite number of them can cover a unit ball.

Addendum:
If we add the hypothesis that all bounded sets are totally bounded (which is equivalent to all balls being totally bounded), then it is true.
The proof hinges on the property of ultrametric spaces that if two open balls are not disjoint, one is a subset of the other; in particular, two
open balls of the same radius are either disjoint or equal.
Proof: For a given open ball $B_0 = B(x, r)$ let $C = \{B(y, r/2) \mid x \in B_0\}$. By the above property, $C$ is a partition of $B_0$. On one hand, $C$ is a maximal cover of $B_0$ by balls of radius $r/2$, since it contains such a ball about every point
of  $B_0$. On the other hand, $C$ is a minimal cover because it is
a partition. Hence $C$ is the only cover of $B_0$ by balls of radius $r/2$ and since total boundedness of $B_0$ implies that there is such a cover that is finite, $C$ must be finite.
Note: If "smaller ball" is interpreted in the stricter sense of "ball that
is a proper subset", then we need the extra assumption that the space
has no isolated points and $C$ can be defined as 
$\{B(y, \delta(B_0)/2 \mid x \in B_0 \}$.
