Does existence of $\omega_1$ subset of reals imply $\omega_1$ choice for subsets of reals? Suppose there exists a subset of $\Bbb R$ which has cardinality $\omega_1$. Is it then necessarilly true that for every collection of $\omega_1$ subsets of $\Bbb R$ there exists a choice function?
I suppose that the answer for the above question is no, just like existence of countable subsets of $\Bbb R$ doesn't imply countable choice for $\Bbb R$. However, I suppose, every proof would either need explicit use of forcing, or a model of ZF which already has some weird properties, like amorphous sets or something.
Note that, obviously, we assume that AC doesn't necessarilly hold.
I added "forcing" tag, because I believe this is what answer requires. If it's not the case, feel free to remove it.
 A: No, since there is a surjection from $\Bbb R$ onto $\omega_1$, there is always an injection from $\omega_1$ into $\mathcal P(\Bbb R)$; but it's not difficult to arrange that there is no choice function for some sequence of subsets like that.
To see that the range of the injection need not have a choice function, simply consider a construction similar to Cohen's first model, only now add $\omega_1$ Cohen reals (and take permutations moving them, etc.). In this model there is a Dedekind-finite set of reals which can be mapped onto $\omega_1$, which defines a sequence of $\omega_1$ sets of reals by taking preimages of each point, but there is no choice function for even countably many of these, since that would show that the set is not Dedekind-finite.

You can in fact get a similar result in Cohen's first model. If $A\subseteq\Bbb R$ is the Dedekind-finite set of generic reals, then $A$ can be mapped onto $\omega$, pick $A_n$ to be the preimage of $n$ under some fixed surjection; and take $A_\alpha=\{x_\alpha\}$ to be the $\alpha$-th real of the ground model ($L$). Then the $A_n$'s don't admit a choice function, so of course the full collection doesn't.
The modification suggested with adding $\omega_1$ Cohen reals can be modified, by taking the filter of subgroups defined by countable supports rather than finite; in the resulting model $\sf DC$ should hold (this should be, I think, $L(\Bbb R)$, which satisfies $\sf DC$). So we have a model that every countable family of sets of reals admits a choice function, but there is a family of size $\aleph_1$ which does not admit a choice function.
