Clearly, it is sufficient to show that any cyclic quadrilateral can be dissected into 5,6 or 7 cyclic quadrilaterals.
Dissection in 5: let $ABCD$ be our initial ciclic quadrilateral, with $X=AC\cap BD$. Let $A',B',C',D'$ the images of $A,B,C,D$ through a circular inversion with respect to a circle centered in X with a small radius. The quadrilateral $A'B'C'D'$ is cyclic since the circumcircle of $ABCD$ is mapped into a circle containing $A',B',C',D'$; $AA'B'B$ is cyclic since $XA\cdot XA'=XB'\cdot XB$.
Dissection in 7: cut from ABCD two small cyclic quadrilateral $ABC'D'$,$A'B'CD$, such that $A'B'$ is parallel to $AB$ and $C'D'$ is parallel to $CD$, then dissect $A'B'C'D'$ into 5 cyclic quadrilaterals.
Dissection in 6: cut from ABCD two small cyclic quadrilateral $ABC'D'$,$A'B'CD$, such that $A'B'$ is parallel to $AB$ and $C'D'$ is parallel to $CD$, then connect the circumcenter of $A'B'C'D'$ with the midpoints of the sides, splitting $A'B'C'D'$ into 4 cyclic quadrilaterals.