Let $X\neq \emptyset$ be a set and let ${\cal A}\subseteq {\cal P}(X)$ with $\bigcup {\cal A}=X$. We say that $S\subseteq X$ is indivisible if for all $T\subseteq S$ with $\emptyset \neq T \neq S$ there is $A\in {\cal A}$ such that $$T \cap A \neq \emptyset \neq (S\setminus T)\cap A.$$ (Note that, vacuously, singletons are indivisible.)

Question. Is there $X\neq \emptyset$ and ${\cal A}\subseteq {\cal P}(X)$ with $\bigcup {\cal A}=X$ and $|A|\geq 2$ for all $A\in {\cal A}$, as well as a cardinal $\kappa>0$, such that the following two statements hold?

1. For every $d:X \to \kappa$ there is $A\in {\cal A}$ such that the restriction $d\restriction_A$ is constant, and
2. Whenever ${\cal S}\subseteq {\cal P}(X)$ is a collection of nonempty, pairwise disjoint, indivisible subsets such that for every $S,T\in {\cal S}$ there is $A\in {\cal A}$ with $A\cap S \neq \emptyset \neq A \cap T$, then $|{\cal S}| < \kappa$.

Let $X\neq \emptyset$ be a set and let ${\cal A}\subseteq {\cal P}(X)$ with $\bigcup {\cal A}=X$. We say that $S\subseteq X$ is indivisible if for all $T\subseteq S$ with $\emptyset \neq T \neq S$ there is $A\in {\cal A}$ such that $$T \cap A \neq \emptyset \neq (S\setminus T)\cap A.$$ (Note that, vacuously, singletons are indivisible.)

Question. Is there $X\neq \emptyset$ and ${\cal A}\subseteq {\cal P}(X)$ with $\bigcup {\cal A}=X$ and $|A|\geq 2$ for all $A\in {\cal A}$, as well as a cardinal $\kappa>0$, such that the following two statements hold?

1. For every $d:X \to \kappa$ there is $A\in {\cal A}$ such that the restriction $d\restriction_A:A \to \kappa$ is constant, and
2. Whenever ${\cal S}\subseteq {\cal P}(X)$ is a collection of nonempty, pairwise disjoint, indivisible subsets such that for every $S,T\in {\cal S}$ there is $A\in {\cal A}$ with $A\cap S \neq \emptyset \neq A \cap T$, then $|{\cal S}| < \kappa$.