Can the following uniformization statement be proved by $ZF+AD+DC$?
For any binary relation $R\subseteq \mathbb{R}^2$ with the property that $\forall x (\{y\mid R(x,y)\}\mbox{ is at most countable and nonempty})$, then there is a function $f:\mathbb{R}\to \mathbb{R}$ uniformizing $R$.
Here's an argument under the further assumption that $V=L(\mathbb{R})$. Something like this is presumably recorded somewhere. The main point comes from Steel's paper "Scales in $L(\mathbb{R})$", but I don't find it quite explicitly there.
Since $V=L(\mathbb{R})$, we can fix a real $x_0$ such that our relation $R$ is $\mathrm{OD}_{x_0}$. Then for each $x\in\mathbb{R}$, $R_x=\{y\bigm|R(x,y)\}$ is $\mathrm{OD}_{(x_0,x)}$. It suffices to see that $R_x$ contains some element which is $\mathrm{OD}_{(x_0,x)}$, since then we can set $f(x)=$ the least such (in the standard ordering of $\mathrm{OD}_{(x_0,x)}$).
In fact, for each real $x$, every countable $\mathrm{OD}_x$ set of reals is $\subseteq\mathrm{OD}_x$. For let's assume $x=\emptyset$; the relativization to other $x$ is routine. Fix a $\Sigma_1$ formula $\varphi$ such that for some ordinals $\alpha<\beta$, the set $$A_{\alpha\beta}=\{y\in\mathbb{R}\bigm|L_\beta(\mathbb{R})\models\varphi(\alpha,\mathbb{R},y)\}$$ is countable; each OD set is of this form. We want to see that each such $A_{\alpha\beta}\subseteq\mathrm{OD}$ (assuming it's countable). Let $A'_{\alpha\beta}=A_{\alpha\beta}$ if $A_{\alpha\beta}$ is countable, and $A'_{\alpha\beta}=\emptyset$ otherwise.
Claim: $A=\bigcup_{\alpha\beta}A'_{\alpha\beta}$ is countable.
Proof: Otherwise we can define a prewellorder $<^*$ on $A$ such that the set of $<^*$-predecessors of any given real is countable, and then argue like in the proof that there is no $\omega_1$-sequence of pairwise distinct reals. QED (Claim).
Note now that $A$ is $\Sigma_1^{L(\mathbb{R})}$ (here the standard notation allows the parameter $\mathbb{R}$ as default, but this is otherwise lightface, i.e. no other parameters). But by Proposition 2.11 of Steel's "Scales in $L(\mathbb{R})$" (applied at some sufficiently large and reflective level $\alpha\in\mathrm{OR}$), and since $A$ is countable, it follows that $A\subseteq\mathrm{OD}$, as desired.
