Prove that for every complex $2$x$2$ matrix $A$, there exists a matrix $X$ such that $X^3=A^2$.

You already have discovered the key point: $A^2$ cannot be a nilpotent matrix of index $2$. This means $A^2$ is either a diagonalisable matrix, or a non-diagonalisable but invertible matrix. The former case is easy. In the latter case, you may (by a similarity transform) assume that $A^2$ is in Jordan form and try to find a triangular matrix cube root of $A^2$.