For each $\alpha<\omega_1$, choose a real $r_\alpha$ that codes the ordertype $\alpha$. This sequence $\langle r_\alpha : \alpha < \omega_1 \rangle$ then codes a sequence of surjections from $\omega$ to each countable ordinal. From this, you can then run the Ulam matrix construction and show that there is no countably complete ultrafilter on $\omega_1$.

Let $F$ be a countably complete filter on $\omega_1$ in $L(\mathbb R^{\omega_1})$, and let $\sigma_\alpha : \omega \to \alpha$ be the surjection coded by $r_\alpha$. For $n<\omega$ and $\beta<\omega_1$, let $S^\beta_n = \{ \alpha > \beta : \sigma_\alpha(n) = \beta \}$. Since $\bigcup_{n<\omega} S_n^\beta$ is the tail interval $(\beta,\omega_1)$, we can take $n_\beta$ to be the least $n$ such that $S^\beta_n$ is $F$-positive, by countable completeness. Let $T_n = \{ \beta : n_\beta = n \}$. Since $\omega_1$ is uncountable, there is $n$ such that $|T_n| > 1$. So for such $T_n$, if $\beta<\gamma$ are in $T_n$ and $\alpha \in S_n^\beta \cap S_n^\gamma$, then $\sigma_\alpha(n) = \beta = \gamma$, which is impossible. Thus $S_n^\beta$ and $S_n^\gamma$ are disjoint $F$-positive sets, and so $F$ is not an ultrafilter.