From the definition of $\beta(r)$ it's obvios that $CC(\beta(r))=CC(r)$.

$CC(\beta(r))=\{A\subseteq U\ |\ \forall X,Y\subset A,\ X\not=\emptyset\not=Y \land X\cup Y=A \land X\cap Y=\emptyset$ $\implies X\cup Y=A\in CC(r)\}=\{A\subseteq U\ | \ A\in CC(r)\}$.

Now assume $A\in CC(\gamma(r))$ and $X,Y\subset A$ non empty such that $X\cup Y=A$ and $X\cap Y=\emptyset$, then $X\gamma(r) Y$. So $\exists N\subset X \land M\subset Y$ such that $NrM$. Now if $r$ is extendable, we have $XrY$. So we can conclude that $A\in CC(r)$.

Is this what you were asking?