Here is the proof of $\omega$-categorical theory having only finitely many formulas on p105 of "Model Theory" (Chang-Keisler 2012). It is part of the Engeler-Ryll-Nardzewski theorem.
Now we assume $(d)$, and prove $(e)$:
For each $n < ∞$, there are only finitely many formulas $\phi(x_1,…,$
$x_n)$ up to equivalence with respect to $T$. Given a formula
$\phi(x_1,… x_n)$, let $\phi^*$ be the set of all types
$\Gamma(x_1,… x_n)$ of $T$ which contain $\phi$. Then
$\phi^*=\varphi^*$ implies $T\models \phi\leftrightarrow \varphi$. But
there are only finitely many types of $T$ in $x_1,…,x_n$, say $m$.
Hence there are only $2^m$ sets of types and therefore at most $2^m$
formulas up to equivalence in $T$.
I am confused on how different types can contain $\phi$ because, by definition, a type $\Gamma(x_1,… x_n)$ is the maximum set of consistent formulas in $x_1,…,x_n$. So formulas in different types should be inconsistent.
Best Answer
(1) "Formulas in different types should be inconsistent". This is wrong. If $p$ and $q$ are distinct types, then there is some formula $\varphi$ such that $\varphi\in p$ and $\lnot \varphi\in q$, and clearly $\varphi$ and $\lnot \varphi$ are inconsistent. This does not mean that for any $\varphi\in p$ and any $\psi\in q$, $\varphi$ and $\psi$ are inconsistent. For example, the valid formulas like $x = x$ will be in every type.
(2) Every formula $\varphi$ determines the set of types $[\varphi] = \{p\mid \varphi\in p\}$. If $[\varphi] = [\psi]$, then $\varphi$ and $\psi$ are equivalent. (Why? if $\varphi$ and $\psi$ were not equivalent, there would be some element $a$ in some model such that, without loss of generality, $a$ satisfies $\varphi$ and $\lnot \psi$. Then $\mathrm{tp}(a)\in [\varphi]$ and $\mathrm{tp}(a)\notin [\psi]$, so $[\varphi]\neq [\psi]$.) So if there are $2^m$ sets of types, there are at most $2^m$ sets of the form $[\varphi]$, so there are at most $2^m$ formulas up to equivalence.