Show that a metric space has uncountably many disjoint dense subsets if every ball is uncountable.

I was wondering how to show that a metric space $X$ has uncountably many dense subsets $Y_{\alpha}$ such that $Y_{\alpha_1} \cap Y_{\alpha_2} = \emptyset$ if $\alpha_1 \neq \alpha_2$, under the assumption that every ball in $X$ is uncountable. When $X = (a,b)$ is a non-empty interval, this is fairly trivial because we can just choose an (uncountable) Hamel basis $H \subset \mathbb{R}$ when ${\mathbb R}$ is a vector space over $\mathbb{Q}$, and then let $Y_\alpha = \alpha {\mathbb Q} \cap X$ for $\alpha \in H$. But what about an arbitrary metric space $X$ where every ball is uncountable?

Claim 1: Let $\kappa$ be an uncountable cardinal and $X$ a metric space in which each open ball has size $\kappa$. Let us call such a space $\kappa$-homogeneous. Then $X$ can be divided into $\kappa$ many dense subsets.

Proof: Clearly $|X| = \kappa$ and hence $X$ has a basis of size $\kappa$ (use rational radii balls), say $\{B_{\alpha} : \alpha < \kappa\}$. Inductively construct $\{D_{\alpha} : \alpha < \kappa\}$ by putting, at stage $\alpha$, one point from each $B_{\beta}$, $\beta < \alpha$ into every $D_{\beta}$, $\beta < \alpha$.

Claim 2: Let $X$ be a metric space in which every open ball is uncountable. Then there is a family of pairwise open balls $U$ of $X$, such that the union of the family is dense in $X$ and each $U$ is a $\kappa$-homogeneous metric space for some uncountable cardinal $\kappa$.

Proof: Every open ball in $X$ contains a homogeneous ball as there is no infinite decreasing sequence of cardinals. So take a maximal family of pairwise disjoint homogeneous open balls in $X$.

Claims 1 + 2 give your result.

The fact that $X$ is a metric space was important here. There are examples of Hausdorff spaces in which every ball is uncountable yet there is no dense codense subset. Such spaces are called irresolvable.