It has been discussed already whether a countable OD set necessarily contains an OD element. See e.g.
A question about ordinal definable real numbers
.
A **negative** answer was obtained in Archive for Mathematical Logic 2015, 54, 5, 711-723, on the base of a very non-homogeneous forcing notion, in fact a clone of Jensen's minimal-$\Pi^1_2$-singleton forcing. In the opposite direction, one would believe that typical homogeneous-forcing models (Cohen, random, Solovay etc) provide a **positive** answer. And indeed it holds in a simple Cohen extension $L[a]$ that

(1) a countable OD set **of reals** necessarily contains an OD element (see arxiv1)

Is the following stronger claim still true in Cohen's $L[a]$?

(2) a countable OD set **of sets of reals** (or weaker but in fact equivalent, an **OD partition of an OD subset $D$ of $\mathbb R$ into countably many pieces**) necessarily contains an OD element

The key step would be to prove (2) restricted to the domain
$$
D=\{x\in\mathbb R\cap L[a]:x\text{ is Cohen and }L[x]=L[a]\}.
$$
And now, one may want to get a counterexample of the following kind. Let $G$ be the group of Borel category-preserving bijections of the reals, coded in $L$. Suppose $G$ is presented in the form $G=C\times U$, where $C$ is a countable subgroup and $U$ a subgroup. Then $D$ is *invariant* under the action of $G$, but one may hope that there are countably many of $U$-orbits in $D$ (as $C$ countable), none of which is OD.

To conclude, is (2) true in Cohen's $L[a]$?

an OD subset of $\mathbb R$into countably many pieces necessarily contains an OD piece $\endgroup$ – Vladimir Kanovei Jul 25 '16 at 6:11