Take $B_r(x)$. Consider $y \notin B_r(x)$. Then $d(y,x)\geq r$. Take $B_r(y)$, then consider a point $z \in B_r(y)$. We'll note $d(y,z)<r$.
Assume that $z \in B_r(x)$, then $d(z,x)<r$.
So we have:
- $d(y,z)<r$
- $d(z,x)<r$
Then $d(y,x) \leq \max\{d(y,z), d(z,x)\}$ implies $d(y,x)<r$, which contradicts earlier established $d(y,x)\geq r$. Hence $z \notin B_r(x), z \in S-B_r(x)$, i.e. $B_r(y) \subseteq S-B_r(x)$, hence $S-B_r(x)$ is open and hence $B_r(x)$ is a closed set $\square$.
Am I correct?
Let $B(x,r)$ be an open ball and suppose $z \in \overline{B(x,r)}$. In particular: $B(z,r)$ intersects $B(x,r)$, say $y \in B(x,r) \cap B(z,r)$. Then $d(x,y) < r$ and $d(y,z) < r$ and so:
$$d(z,x) \le \max(d(z,y), d(y,x)) < r$$
or $z \in B(x,r)$ and so $B(x,r)$ is closed, as it already contains every point of its closure.
Another fun fact: $B(x,r) = B(y,r)$ for every point $y \in B(x,r)$, which I remember as "every point of an ultrametric ball is its centre". The proof is similar to the above one.