RNP is separably determined

Transfer dentability to a closed subspace. Assume $X$ has RNP and let $Y\subseteq X$ be a closed separable subspace. Any nonempty bounded closed convex $C\subseteq Y$ is also closed in $X$. By [L2] it has arbitrarily small slices determined by functionals in $X^{\ast }$. Restricting such a functional to $Y$ gives the same slice of $C$, including the zero-functional singleton case. Thus every such $C$ is dentable in $Y$.

Conclude the forward implication. Apply the reverse direction of [L2] inside $Y$ to conclude that $Y$ has RNP. Hence RNP passes to every closed separable subspace.

Obtain the separable-range witness for the converse. Now assume every closed separable subspace of $X$ has RNP. If $X$ failed RNP, [L2] would give a nondentable bounded closed convex set, and [L3] would yield an absolutely continuous bounded-variation vector measure $\nu$ on $[0,1]$ with no $X$-valued density and with range in a closed separable subspace $Y$.

Use the subspace density to contradict the witness. Regarded as a $Y$-valued measure, $\nu$ has the same variation and absolute continuity. By the assumed RNP of $Y$ and [L1], it has a Bochner density $h:[0,1]\to Y$. The isometric inclusion $i:Y\hookrightarrow X$ is bounded; [L4] gives $\nu (E)=i({\int }_{E}h)={\int }_{E}i\circ h$ in $X$, contradicting step 3.1.

Combine both directions and record boundaries. [A1, step 2.1, step 4.1] Steps 2.1 and 4.1 prove the equivalence. The zero subspace is closed and separable and has its zero density; if $X=\{0\}$ both sides hold. The full-AC cost is precisely that inherited from [L2] and [L3].