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Aug 28 at 4:42 comment added Elliot Glazer @GabeGoldberg Assume $V=L,$ let $\kappa$ be inaccessible, let $G=\langle G_{\alpha} : \alpha < \kappa \rangle$ be Levy collapse generic. Let $R=\mathbb{R}^{L[G]}$ and define $f: R \rightarrow \kappa$ by setting $f(x)$ to be the least $\alpha$ such that $x \in L[G \restriction \alpha].$ Then $M=L(R, f)$ has $(R, \prec)$ as in the question, and every set of ordinals in $M$ is OD from $f$ and some $G \restriction \alpha,$ but $R \cap HOD^M(f, G \restriction \alpha) = R \cap L[G \restriction \alpha]$ is countable in $M.$ So no subset of $\omega_1$ codes an injection from $\omega_1$ to $R.$
May 29 at 14:17 comment added Asaf Karagila Oh, that's an interesting question. I don't know.
May 28 at 0:48 vote accept Ollie
May 27 at 23:29 comment added Gabe Goldberg What about the case where such an order exists on $\mathbb R$ itself?
May 27 at 22:20 history answered Asaf Karagila CC BY-SA 4.0