Let $D$ be the set of all recursive (aka decidable) sets of integers, and $C$ be the set of all constructible sets of integers , i.e. $C=L\cap {\cal P}(\omega)$ where $L$ is Godel's constructible universe. What inclusions hold or do not hold between $C$ and $D$ ?
My thoughts : $D$ is countable since there are only countably many programs/algorithms to decide the sets. On the other hand, $C$ is uncountable inside $L$ because $L$ is a model of $ZFC$, and my guess is that $C$ should stay uncountable inside $V$, so the inclusion $C \subseteq D$ should be false.
You are right. $L$ contains all Turing machines of $V$ (since Turing machines are/can be coded as elements of $V_{\omega} = V_{\omega}^{L}$) and it evaluates Turing machines correctly. So $D = D^L$, where $D^L$ is the set of all subset of $\omega$ that are decidable from the point of view of $L$. On the other hand $$ C = \mathcal P(\omega) \cap L = \mathcal P(\omega)^L, $$ i.e. $C$ is the powerset of $\omega$ as constructed in $L$. Since $\operatorname{ZFC}$ proves that there are non-recursive subsets of $\omega$, we have that $$ L \models D^L \subsetneq \mathcal P(\omega), $$ i.e. $D = D^L \subsetneq C$.
We can be much more precise than that: Let $$ E = \{ I \in \omega \mid I = \langle M \rangle \text{ for some Turing machine } M \text{ that holds on empty input} \}, $$ where $\langle . \rangle$ is some primitive recursive coding. Then $E = E^L \in C$ and $E \not \in D$ (this is the halting problem). So $E$ is a canonical witness for $D \neq C$.