I've studied the proof of the independence of $\text{AC}_\omega (\mathbb{R})$ from $\text{ZF}$ using symmetric models, but I was wondering how to prove the independence of $\text{AC}_\omega (\mathbb{R})$ from $\text{ZF}+\text{V}=\text{L}(\mathbb{R})$.
Could I simply add $\omega$ Cohen reals to $\text{L}$ and then take Cohen's symmetric submodel of the generic extension? Would the symmetric model satisfy $\text{V}=\text{L}(\mathbb{R})$?
Thanks
Cohen's model is not of the form $L(\Bbb R)$, to see that note that the Dedekind-finite set of the generic reals is not constructible from any single real.
Moreover, if you do take $L(\Bbb R)$ of the Cohen model, you get Feferman's model in which $\omega$ carries no free ultrafilters. And funny enough, that model satisfies $\sf DC$...
The only proof I am familiar with is the one of Truss, where $\omega_1$ is singular. Namely, collapse a singular cardinal to be $\omega_1$, take everything definable from bounded sequences of collapses, so you get a Solovay model, but we started with a singular cardinal. That model, assuming we started with $L$ as our ground model, will be $L(\Bbb R)$, and it will recognise that $\omega_1$ is singular, hence the failure of $\sf AC_\omega(\Bbb R)$.