Consider the process $X(t) = A(t) + M(t)$, where $\{A\}$ is a finite variation predictable process and $\{M\}$ is a local martingale.
Suppose that $\{A\}$ does not have any jump, hence it is a continuous finite variation process. The quadratic variation of such process is $0$.
Is it right to conclude that $[X,X]_t$ = $[M,M]_t$?
Yes. Cauchy-Schwarz applies to the quadratic variation (you can see this from the definition of the QV), so
$$[A,M]_t^2 \le [A,A]_t[M,M]_t = 0,$$ and
$$[X,X]_t = [A,A]_t + 2[A,M]_t + [M,M]_t = [M,M]_t.$$