Lemma 2.9 if $(X,d)$ is an ultrametric space, then any two open balls of radius $r>0$ are eithier equal or disjiont. In particular, if X is compact, then every open ball of radius r is clopen and the collection of all open balls of radius r forms a partition.
let $B_r(x) $ and $B'_r(y)$ be open balls that exist assume $z \in B_r(x)\cap B'_r(y) $ then $d(x,y) < max \{ d(x,z),d(z,y)\} < r $ hence $d(x,y) < r $ and so $x,y$ are in the same ball. now additionally for any $x_i \in B_r(x)$ the same argument shows that it is in $B'_r(y) $ it follows that if the intersection is non-empty then the balls are the same. if it is empty then they are disjiont by definition as both are openz.
Now we show that they cover the whole space; let $r>0$ then $ \cup_{\{x\in X \}}B_r(x)=X$ since X is compact and theses balls cover the space there is a finite sub-cover that covers the space.
Now we have a finite union of open balls that cover $X $ that are disjoint hence a partition.
Now i have nfi how to show that each ball is also closed.
Best Answer
Each ball $B_r(x)$ is open; this holds on any metric space. But you proved that $X\setminus B_r(x)$ is the union of finitely many open balls. So, $X\setminus B_r(x)$ is open and therefore $B_r(x)$ is closed.