"Reverse Resolutions" exist for $\aleph_0$ categorical theories. Suppose $T$ is a countable $\aleph_0$ categorical theory and that $M \vDash T$ is countable. Then there exists a countable descending elementary chain $\{M_n : n < \omega\}$ such that $M_{n+1} \prec M_n \forall n$, each $M_n$ is countable, and moreover $M = \cap M_n$. (This is a proper chain of course, that is, we can't just take $M_n = M$).
I would appreciate help for this. I don't think we're expected to use anything beyond types machinery for this, but I'm having no luck.
 A: First note that the assumption of $\aleph_0$-categoricity allows us to reduce the problem to the following one:

Find a sequence of countable models $(M_n)_{n\in \omega}$ such that $M_{n+1}\prec M_n$ for all $n$, and $M_\infty = \bigcap_{n\in \omega} M_n$ is a countable model of $T$.

Indeed, if $M\models T$ is any countable model, then $M\cong M_\infty$ (since $T$ is $\aleph_0$-categorical), so by replacing the elements of $M_\infty$ in all the $M_n$ by the corresponding elements of $M$, we can arrange that $\bigcap_{n\in \omega} M_n = M$.
This reduction is the only place we'll use the $\aleph_0$-categoricity assumption. The statement is actually true for an arbitrary (infinite) model $M$ of an arbitrary theory $T$, but realizing an arbitrary model as an intersection of a descending elementary chain is more difficult (see my answer here).
Ok, now let's achieve our goal.
Step 1: Skolemize. Let $L'$ be a countable language expanding $L$ by Skolem functions, and let $T'$ be the Skolemization of $T$ in the language $L'$. The point of doing this is that for any model $N\models T'$, every substructure of $N$ is in fact an elementary substructure and hence a model of $T'$. In particular, the intersection of any family of substructures of $N$ is a model of $T'$.
So all we need to do is find a sequence of countable models $(M_n)_{n\in \omega}$ of $T'$ such that $M_{n+1}$ is a proper substructure of $M_n$ for all $n$. The intersection $M_\infty$ will automatically be a countable model of $T'$. If we don't Skolemize, the intersection of the descending elementary chain might be empty!
Note that $T'$ is no longer $\aleph_0$-categorical. This is ok - as I promised above, we only need the $\aleph_0$-categoricity assumption for the reduction.
Step 2: Find the descending sequence. Consider the partial type $\Delta$ in the variables $(x_n)_{n\in \omega}$ given by $$\{x_n\neq t\mid t\text{ is an $L'$-term with free variables from }(x_k)_{k > n}\}.$$
Show that $\Delta$ is consistent by compactness, and realize it by a sequence $(a_n)_{n\in \omega}$ in a model $N\models T$. Now for all $n$, let $M_n$ be the substructure of $N$ generated by $\{a_k\mid k\geq n\}$. Note that $M_n$ is countable, since it is generated by a countable set in a countable language. Clearly $M_{n+1}\subseteq M_n$, since it is generated by a smaller set, and moreover $M_{n+1}$ is a proper substructure of $M_n$, since $a_n\in M_n\setminus M_{n+1}$.
Step 3: By the above construction, we have a sequence of countable models of $T'$ such that each is a proper elementary substructure of the previous, and their intersection is a countable model of $T'$. It remains to take the reduct back to $L$ and note that the same conditions hold when we view the $M_n$ as models of $T$.
