Let $L$ be a countable first order language and let $T$ be a theory in $L$ with countably many non-isomorphic countable models, then there is a $L$-sentence $\phi$ such that the theory axiomatized by $T\cup\{\phi\}$ is complete.
Here is the skeleton of my proof which may or may not work:
If $T$ is complete, let $\phi\in T$, then $T\cup\{\phi\}=T$ is complete and we're done.
Otherwise if $T$ is not complete. Let $\phi$ be such that $\phi\notin T$ and $\neg\phi\notin T$.
By Vaught's theorem, it suffices to show: 1. $Ent(T\cup\{\phi\})$ has no finite model, and 2. there exists a cardinal $\kappa\geq\aleph_0+|L|=\aleph_0$ (since $L$ is countable) such that if $M\vDash Ent(T\cup\{\phi\})$ with $|M|=\kappa$, then $M$ is unique up to isomorphism.
Since $T$ is a theory, we have $T=Ent(T)$, where $Ent(T):=\{\phi\mid$ if $M\vDash T$ then $M\vDash\phi\}$.
For 1: Suppose for a contradiction $T\cup\{\phi\}$ has a finite model $M$. Since $\phi\notin Ent(T)=T$, there is a model $M'$ such that $M'\vDash T$ and $M'\vDash\neg\phi$. How can I find a contradiction from this?
For 2, let $M_1$ and $M_2$ be models of $Ent(T\cup\{\phi\})$ such that $|M_1|=|M_2|\geq\aleph_0$. We must show that $M_1\cong M_2$. But I'm not sure how to proceed?
Where should I use the countability assumptions?
Hint. Show that, if no finite extension of $T$ is complete, then $T$ has $2^{\aleph_0}$ complete extensions. Each of those extensions has a countable model, and no two of those models are elementarily equivalent, much less isomorphic.