I'm trying to understand the definition of a cantor-bendixson rank for a formula. It is defined as (in my notation) for $\phi(x)$ in fixed $L$ and complete theory $T$,
(i)$CB(\phi(x))\geq 0$ if $\phi$ is consistent with $T$
(ii)$CB(\phi(x))\geq\alpha$($\alpha$ is limit ordinal) if $CB(\phi(x))\geq\beta$ for all $\beta<\alpha$
(iii)$CB(\phi(x))\geq\alpha +1$ if there's a pairwise inconsistent $(\phi_i |i<\omega)$ such that $CB(\phi(x) \wedge \phi_i (x)\geq\alpha$
I think CB-rank measures a kind of degree of nonisolatedness of $[\phi]=\{p\in S_n (T) | \phi\in p\}$ where $S_n(T)$ is a stone space consisting complete types of $T$. I'm tryng to prove if $\phi$ does not isolate any $p\in S_n(T)$, then $CB(\phi)\geq 1$.
I thought since $\phi$ does not isolate any complete types, $|[\phi]|\geq\omega$ since stone space is hausdorff. then I thought I have to pick pairwise inconsistent formulas from different types in $[\phi]$. but I couldn't proceed anymore.
What you're trying to prove is not true. Fix a language $L$ with two constant symbols $c$ and $d$ and any complete theory $T$ which proves $c\neq d$. Let $\varphi(x)$ be the formula $(x = c)\lor (x = d)$. Then $\varphi(x)$ has CB-rank $0$, but it does not isolate any type: there are two types consistent with $\varphi(x)$, namely those isolated by $x = c$ and $x = d$.
This also provides a counterexample to your assertion that if $\varphi$ does not isolate a complete type, then $|[\varphi]|\geq \omega$, since we have $|[\varphi]| = 2$.
The correct statement is that the following are equivalent: