Here is a problem in "Model Theory" (Exercise 2.3.20, Chang-Keisler 2012).
In a countable language, let $T$ be a complete theory which is not $\omega$-categorical. Let $\Gamma_1, \cdots, \Gamma_n$ be consistent types over $T$. Show that $T$ has a countable model $\frak{A}$ which realizes $\Gamma_1, \cdots, \Gamma_n$ but is not saturated.
To prove it, it needs to show that a finite type can not be realized in $T$. But since each $\Gamma_i$ is realized in $T$, I am not sure how to do it.
Let $\mathcal U$ be a saturated model of $T$.
Fact. The following are equivalent for $a_1,\dots,a_n\in\mathcal U$ some arbitrary elements
$T$ is $\omega$-categorical
every elementary substructure of $\mathcal U$ is $\omega$-saturated
every elementary substructure of $\mathcal U$ containing $a_1,\dots,a_n$ is $\omega$-saturated.
Assume the Fact (exercise).
Let $a_1,\dots,a_n$ be realizations of $\Gamma_1,\dots,\Gamma_n$ in $\mathcal U$. If $T$ is not $\omega$-categorical, there is an elementary substructure $\frak A$ of $\mathcal U$ containing $a_1,\dots,a_n$ which is not saturated.
Proof of the Fact
(0$\Rightarrow$1) Negate $(1)$ and assume that $\frak A\preceq\mathcal U$ does not realize the consistent type $\Gamma$. By Löwenheim-Skolem we may assume $\frak A$ is countable. Again by Löwenheim-Skolem there is a countable $\frak B\preceq\mathcal U$ that realizes $\Gamma$. As $\frak A, B$ are not isomorphic, $T$ is not $\omega$-categorical.
(1$\Rightarrow$2) Clear.
(2$\Rightarrow$0) Any two countable $\frak A, B$ such that $a_1,\dots,a_n\in\frak A, B\preceq\mathcal U$ are saturated. Hence they are isomorphic over $a_1,\dots,a_n$. Then the theory Th$({\mathcal U}/a_1,\dots,a_n)$ is $\omega$-categorical. Then for every $k$ there are finitely many non equivalent formulas $\varphi(a_1,\dots,a_n,x_1,\dots,x_k)$. In particular there are also finitely many non equivalent formulas $\psi(x_1,\dots,x_k)$.