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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.

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. In other words, if $S$ is a set of pairwise disjoint nonempty sets, and if $|\bigcup S|\le2^{\aleph_\alpha}$ for some ordinal $\alpha$, then $S$ has a choice function.

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.

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.

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 $f(a)\in a$.

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.