It seems to me that your **DpC** implies that every set of pairwise disjoint nonempty subsets of $\mathbb R$ has a choice function; equivalently, that every set of pairwise disjoint nonempty subsets of $[\omega]^\omega$ has a choice function.

Let $S$ be a set of pairwise disjoint nonempty subsets of $[\omega]^\omega$. Let $X=S\cup\{\{n\}:n\in\omega\}$, still a set of pairwise disjoint nonempty sets. Let $f$ be an injective "deep choice function" for $X$. Then $f(\{n\})=n$ for each $n\in\omega$. If $a\in S$, then $\operatorname{tc}(a)\subseteq a\cup\omega$, so $f(a)\in a$.

**P.S.** I guess "every set of pairwise disjoint nonempty subsets of $\mathbb R$ has a choice function" implies that "every set of at most $\mathfrak c$ nonempty subsets of $\mathbb R$ has a choice function." I don't know about the converse.

**P.P.S.** I guess a similar argument shows that every set of pairwise disjoint nonempty subsets of $[\omega_\alpha]^{\omega_\alpha}$ has a choice function.