Let $(X, d)$ be a topological space with ultrametric $d$. Under what assumption can we say that every open ball $B(a, r) = \{x \in X : d(a, x) < r\}$ is also a closed ball? This is certainly true for valued fields $X = K$ if their valuation group $G = |K|$ has only one accomodation point, zero (example: $K = \mathbb Q_p$, the $p$-adic numers, where $B(a, r) = B[a, r - \varepsilon]$ for small values of $\varepsilon$).
I already know that every open ball is also a closed set in this setting.
I think the phenomenon must be much more general than for the case of discrete value groups. Let’s show it for any ultrametric field $K$, with absolute value $|\bullet|$.
I’ll show that the open ball $B(0,r)$ is also closed; this should do it. Indeed, suppose that $z\notin B(0,r)$, so that $|z|\ge r$. Consider $B(z,r/2)$, an open neighborhood of $z$ that’s disjoint from $B(0,r)$, because $|z-w|<r/2\Rightarrow|w|=|(w-z)+z|=\max(|w-z|,|z|)=|z|\ge r$. So every element of the complement of $B(0,r)$ has an open neighborhood that’s also disjoint from $B(0,r)$. Thus that complement is open, and $B(0,r)$ is closed.
EDIT:
As I say in my comment, the above has failed to address the most important aspect of your question, and I apologize.
It seems to me that if the value group of your field is infinite and not cyclic, then the unit open ball will necessarily fail to be a closed ball. Under what circumstances a valued field $K$ will have open balls that are closed seems to me more subtle, and I’ll have to leave that to specialists. But in the case of $\Bbb C_p$, the completion of the (an) algebraic closure of $\Bbb Q_p$, then since its value group is $p^{\Bbb Q}$, there are balls that are simultaneously open balls and closed balls, such as $\{z\in\Bbb C_p:|z|<p^{\sqrt2}\}$.