$\sf LS(\aleph_0)$: Every model $M$ of a first order theory $\sf T$ with countable signature has an elementary submodel $N$ which is at most countable.
Now this theorem is equivalent over axioms of $\sf ZF$ to the axiom of dependet choice $“\sf DC"$.
Now does this mean that in $\sf ZF + \neg DC$, we can have a consistent theorey in a countable signature that doesn't have a countable model?
Best Answer
No. If a theory is well-orderable, then it has a well-orderable model satisfying all the usual properties in $\sf ZFC$.
To see why, note that you can code all the thing into a set of ordinals $A$, then in $L[A]$ your theory exists and it is a model of $\sf ZFC$, so the theory has a countable model. But this is upwards absolute to $V$.