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Under AD, one can have DC plus there is no $\omega_1$-sequence of distinct reals. So this is a case where one can have DC, plus $\omega_1\not\leq \mathbb{R}$, but as you noted, there is always a surjection from $\mathbb{R}$ onto $\omega_1$, since every countable ordinal is coded by a real, and so $\omega_1\leq^\ast\mathbb{R}$. So this is a case where $H(\mathbb{R})=\omega_1\leq^\ast\mathbb{R}$ and DC holds. (I don't think you need full AD to get this situation.)

Do you want more choice than DC? For which choice principle specifically are you aiming?

Perhaps one should look at the model $L(V_{\lambda+1})$, the higher analogue of $L(\mathbb{R})$, to get a greater degree of choice, but by analogy one might still expect that $H(V_{\lambda+1})=\lambda\leq^\ast H(V_{\lambda+1})=\lambda^+\leq^\ast V_{\lambda+1}$.

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Under AD, one can have DC plus there is no $\omega_1$-sequence of distinct reals. So this is a case where one can have DC, plus $\omega_1\not\leq \mathbb{R}$, but as you noted, there is always a surjection from $\mathbb{R}$ onto $\omega_1$, since every countable ordinal is coded by a real, and so $\omega_1\leq^\ast\mathbb{R}$. So this is a case where $H(\mathbb{R})=\omega_1\leq^\ast\mathbb{R}$ and DC holds. (I don't think you need full AD to get this situation.)

Do you want more choice than DC? For which choice principle specifically are you aiming?

Perhaps one should look at the model $L(V_{\lambda+1})$, the higher analogue of $L(\mathbb{R})$, to get a greater degree of choice, but by analogy one might still expect that $H(V_{\lambda+1})=\lambda\leq^\ast V_{\lambda+1}$.