Let $R$ be a valuation ring, with fraction field $K$ and residue field $k$. Denote by $\Gamma=K^{\times}/R^{\times}$ the valuation group (assumed nontrivial).

The valuation $v:K^{\times}\to\Gamma$ decomposes $K^{\times}$ as a disjoint union of nonempty open subsets, indexed by $\Gamma$. Each of these is homeomorphic to $R^{\times}$, which is in turn (using the reduction map to $k$) a disjoint union of nonempty open subsets, indexed by $k^{\times}$.

We conclude that any basis for the topology of $K$ must have cardinality at least $\kappa:=\max(\mathrm{Card}\,\Gamma, \mathrm{Card}\,k)$.

Question: does there exist a basis of open subets of $K$ with cardinality $\kappa$?

Remarks:

(1) It is true if $v$ is discrete, i.e. $\Gamma\cong\mathbb{Z}$. Proof: take a set $S\subset R$ of representatives of $k$, and a uniformizing parameter $\pi$. Let $X\subset K$ be the set of finite sums $\sum_i s_i\pi^{n_i}$ ($s_i\in S$, $n_i\in\mathbb{Z}$). Then the balls centered on $X$ form a basis.

(2) I am especially intereseted in the case $\kappa=\omega$. Explicitly: if $\Gamma$ and $k$ are countable, does it follow that $K$ is second-countable (or, equivalently, separable)?