If $0^\sharp$ exists then the $L$-indiscernibles form a proper class of ordinals without any infinite constructible subset, as $0^\sharp$ can be defined from any infinite increasing sequence $\langle \kappa_i\mid i<\omega\rangle$ of them as $$0^\sharp = \{\varphi(v_0,\dots, v_n)\mid L_{\sup_{i<\omega}\kappa_i}\models\varphi(\kappa_0,\dots, \kappa_n)\}$$ On the other hand we, can force a set of ordinals of arbitrary size $\lambda$ without infinite constructible subsets by forcing with finite partial functions $p:\lambda\rightarrow 2$ (which is the same as adding $\lambda$-many Cohen reals). But clearly the class-sized version of this does not preserve $\mathrm{ZFC}$.
My question is:
Is there a model of $\mathrm{ZFC}$ with a definable proper class of ordinals without an infinite constructible subset, yet $0^\sharp$ does not exist in this model?
Such a model cannot be a generic extension (by set-sized forcing) of $L$, as otherwise one condition in the generic filter must force an infinite amount of ordinals into this class and $L$ can see this.
This question is related to this question by Joel Hamkins in the following way: Suppose $V$ has a nonconstructible real and there is a definable embedding $\pi:(V, \in)\rightarrow (L, \in)$ (in the sense of that question). Let $C$ be the range of $\mu\circ\pi$ where $\mu$ sends a set in $L$ to its rank in the canonical global wellorder. Then by the answer of Joel Hamkins, $0^\sharp$ does not exist and the arguments in the answer of Farmer Schlutzenberg show that $C$ does not have an infinite constructible subset. It seems to be open whether this scenario is consistent.
(I have asked this question on MSE before but it did not get answered there)
