Is every type realized in a countable model?

Question

Let $p(x)$ be a type of a theory $T$ over the empty set. Is $p$ realized in a countable model of $T$?

I think this is the case, because if add a symbol $\mathring x$ to the language $L$, then the set of formulas $T \cup p(\mathring x)$ is consistent, so it has model $M$, then by Lowenheim-Skolem there is some countable elementary $(L+\mathring x)$-substructure $N$ of $M$ and the reduct of $N$ to $L$ is a model of $T$ realizing $p$. Is my reasoning correct?

Answer 1

Your argument is correct.

It also makes a nice proof by contradiction.

Suppose for contradiction that the 1-type without parameters $p(x)$ was not satisfiable in any countable model,

then $T \cup p(c)$ with $c$ being a fresh constant symbol would be satisfiable in some infinite structure but not in any countably infinite structures.

This contradicts the Löwenheim–Skolem theorem.