I found this theorem in T. Tao's notes on totally transcendental theories. A theory is t.t. if there is no binary tree of $L(M)$-formulas for any model $M$. I don't understand something about the proof. I'll state the theorem, and after I'll insert an image of the proof.
Theorem 1.4. In a countable language $L$, consider a complete $L$-theory $T$. TFAE:
- $T$ is t.t.;
- $T$ is $\lambda$-stable, for any $\lambda\geq\aleph_0$;
- $T$ is $\omega$-stable.
I don't understand why he can say that, "even if we exclude all small formulas from the set $U_{\phi}$, its cardinality is still greater than $\lambda$".
Your question is really about cardinal arithmetic, not t.t. theories.
If $\psi(\overline{x})$ is small, then $|U_\psi|\leq \lambda$. Now there are only $\lambda$-many $L(A)$-formulas, so there are only $\lambda$-many small $L(A)$-formulas. Thus $$\bigcup_{\psi(\overline{x}) \text{ small}}U_\psi$$ is a union of at most $\lambda$-many sets, each of size at most $\lambda$, so $$\left|\bigcup_{\psi(\overline{x}) \text{ small}}U_\psi\right| \leq \lambda\times \lambda = \lambda.$$
But since $U_\phi$ is large, $|U_\phi|>\lambda$. It follows that $$\left|U_\phi\setminus \bigcup_{\psi(\overline{x}) \text{ small}}U_\psi\right| = |U_\phi| >\lambda.$$
Why? Well, if the cardinality of this set were $\leq \lambda$, then we could write $U_\phi$ as a disjoint union of two sets of size $\leq \lambda$, so we would have $|U_\phi| \leq 2\times \lambda = \lambda$, which is a contradiction.
In particular, the cardinality of this set is $\geq 2$, which is the conclusion that is used in the proof.