Is there always a matrix $X$ such that $X^2=A$? 
Is it true that for every $A\in M_{2\times 2} (\mathbb{C})$ there's an $X\in M_{2\times 2} (\mathbb{C})$ such that $X^2=A$?

For the matter of fact, I don't have a clue, other than evaluating the general case and solve the system of equations.
I also thought about the fact that we're aobve $\mathbb{C}$ and therefore the characteristic polynomial can be factored to linear terms.

No: the matrix $E=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is such that $E^2=0$. If there were a matrix $X$ such that $X^2=E$, it would be nilpotent. However, by Hamilton-Cayley, the index of nilpotency (the smallest $k$ such that $X^k=0$) is at most $2$ in a space of dimension $2$, so that we would have $X^2=0=E$.
