Maximally consistent theories have complete countable subtheories in every countable sublanguage. Given a maximally consistent theory $T$ in a language $L$, show that, for every countable $L_0 \subseteq L$, there is a $T_0 \subseteq T$ that is complete.
I have an attempt at a solution but it doesn't seem to use the countable-ness condition, and seems too easy.
Suppose $L_0 \subseteq L$. Let $T_0 := \{\varphi \in T : \varphi \in L_0\}$. Then for every $\varphi \in L_0$, either $T \cup \varphi$ or $T \cup \neg \varphi$ is consistent, which implies $\varphi \in T$ or $\neg \varphi \in T$ which means $\varphi \in T_0$ or $\neg \varphi \in T_0$ and thus $T \vdash \varphi$ or $T \vdash \neg \varphi$.
This seems too easy however, and it does not use the fact that $L_0$ is countable. Can someone check my proof please? Thank you!
 A: Yes, the problem as phrased is exactly as simple as it appears. Indeed, you've proved more (as you've observed):

If $T$ is a maximally consistent $L$-theory then for every sublanguage $L_0\subseteq L$ the subtheory $T\cap Sent(L_0)$ is a maximally consistent $L_0$-theory.

It's worth noting that there is, however, a subtlety which crops up re: countable vs. uncountable languages with respect to satisfiability. Namely, the statement "every consistent theory in a countable language has a model" is provable in set theory without the axiom of choice (and indeed we can replace "countable" with "well-orderable"), but the full completeness theorem is not. So the following is a nontrivial theorem:

$(\mathsf{ZF}$): If $T$ is a consistent $L$-theory and $L_0$ is a countable (or indeed well-orderable) sublanguage of $L$, then $T\cap Sent(L_0)$ is satisfiable. Moreover, if $T$ is maximally consistent then there is a structure $\mathcal{M}_0$ such that $T\cap Sent(L_0)=Th(\mathcal{M}_0)$.

But it sounds like that's not relevant to what you're doing. I suspect that your instructor made a typo, or had some other kind of silly moment.
