$C^-$ is denoted $C^+$. The fact that $A$ has maximal rank is useless.
1) We show that, for every $X,C$, one has $rank(X(I_p-C^+C))=dim(\ker(C))-dim(\ker(X)\cap \ker(C))$.
Proof: $C^+C$ is the orthogonal projection onto $im(C^*)$, that is, onto the orthogonal of $\ker(C)$. Then $I_p-C^+C$ is the orthogonal projection onto $\ker(C)$. Thus $im(I-C^+C)=\ker(C)$ and $im(X(I-C^+C))=X(\ker(C))$. Let $\ker(C)=(\ker(C)\cap\ker(X))\oplus Z$ ; then $X(\ker(C))=X(Z)$ and $X$ is an isomorphism from $Z$ onto $X(\ker(C))$ and we are done.
2) Here $\ker(X)\subset \ker(C)$ ; we deduce that $rank(X(I_p-C^+C))=dim(\ker(C))-dim(\ker(X))=rank(X)-rank(C)$.
EDIT: @Dunn,@Gigili, I see that you are unable to send a simple "thank you". Incredible rudeness.