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)*