Suppose that there is an $n$ such that the set of $L$-formulas in variables $x_1,\dots,x_n$ up to $T$-equivalence is finite. That is, there is no finite set of formulas $\{\varphi_i\}_{i\in I}$ (with $I$ finite) such that for all $L$-formulas $\psi(x_1, x_2, \ldots, x_n)$ there is a $i \in I$ such that $$T \models \forall x_1, x_2, \ldots, x_n(\psi(x_1, x_2, \ldots, x_n) \leftrightarrow \varphi_i(x_1, x_2, \ldots, x_n)).$$
Then somehow by the the topological compactness of $S_n(T)$, there must be an non isolated type $p(x)$. My question is how to conclude that from the hypothesis.
I understand that if $S_n(T)$ is compact then for an open cover $\mathcal{C}$ of $S_n(T)$ there is a finite cover $\mathcal{C}'$ of $S_n(T)$, that is if $S_n(T) = \bigcup_{[\varphi] \in \mathcal{C}} [\varphi]$, then $S_n(T)= \bigcup_{[\varphi] \in \mathcal{C}'} [\varphi]$, but from this information I do not know how to obtain a non isolated type $p(x)\in S_n(T)$.
Is there any other property of the compactness of $S_n(T)$ that I am missing? Can you explicitly find the non isolated type, or is the proof by contradiction, i.e. supposing that all types of $S_n(T)$ are isolated and using the fact that the set $\{\varphi_i\}_{i\in I}$ is infinite?
Let's prove the contrapositive. Suppose every type in $S_n(T)$ is isolated. Then for all $p\in S_n(T)$, $\{p\}$ is open, and $\bigcup_{p\in S_n(T)}\{p\}$ is an open cover of $S_n(T)$. By compactness, $S_n(T)$ is finite.
Now we can write $S_n(T) = \{p_1,\dots,p_k\}$, and since each $p_i$ is isolated, we can find a formula $\varphi_i(x_1,\dots,x_n)$ such that $[\varphi_i] = \{p_i\}$. For each $S\subseteq \{1,\dots,k\}$, let $\varphi_S$ be $\bigvee_{i\in S} \varphi_i$. Then $[\varphi_S] = \{p_i\mid i\in S\}$.
Finally, for any formula $\psi(x_1,\dots,x_n)$, let $S_\psi = \{i\mid p_i\in [\psi]\}\subseteq \{1,\dots,k\}$. Then $[\psi] = [\varphi_{S_\psi}]$, so $$T\models \forall x_1\dots x_n (\psi(x_1,\dots,x_n)\leftrightarrow \varphi_{S_\psi}(x_1,\dots,x_n)).$$ Since there are only finitely many formulas of the form $\varphi_S$, this shows that there are only finitely many formulas up to $T$-equivalence.