Here is an argument/explanation that does not use Jordan normal form or charateristic polynomials. Instead it uses a more linear transformation-based viewpoint, and properties of projectors. I find this a useful way to think about these kinds of questions, which is why I'm positing it.
Note first that $A^3 = A$ implies that $A^4 = A^2$. This means that $A^2$ is a projection. In general, if $P^2 = P$, for some linear transformation of a vector space $V$ (over any field), then $V$ is the direct sum of the image of $P$ and the image of $I- P$. (Here $I$ denotes the identity.) Furthermore, on the the image of $P$, the transformation $P$ acts by the identity, while on the image of $I - P$, it acts by zero.
So in our case, the $3$-dim'l vector space $V$ on which your matrix $A$ is acting splits as the direct sum of the image of $A^2$, on which $A^2$ acts by the identity, and as the image of $I - A^2$, on which (using the equation $A - A^3 = 0$) we see that $A$ acts by zero.
So we have partially diagonalized $A$; we have decomposed $V$ into a sum of two subspaces, each invariant under $A$, with $A^2 = I$ on the first, and $A = 0$ on the second.
To complete the diagonalization, we make the same kind of argment, but now we may assume that $A^2 = I$.
From $A^2 = I$, we find that $\bigl(\dfrac{I-A}{2}\bigr)^2 = \dfrac{I-A}{2}$. Thus $(I-A)/2$ is again a projector, and so the subspace on which $A^2 = I$ decomposes as a sum of two subspaces, one on which $(I-A)/2 = I,$ which is to say $A = - I$, and one on which $(I-A)/2 = 0$, which is to say $A = I$.
Putting it altogether, we decomposed our original space $V$ as the sum of three $A$-invariant subspaces, on which $A$ acts by $0$, $-1$, and $1$ respectively.
The argument works over any field where $2$ is invertible. If we are in char. $2$, then the decomposition into the sum of spaces on which $A = 0$ and $A^2 = I$ is still possible, but (as Joriki points out in his answer) we can't necessarily diagonalize a matrix satisfying $A^2 = I$.
One way to see this is to note that in char. $2$, $A^2 = I$ is equivalent to $(A-I)^2 = 0$, and so we can construct matrices $A$ such that $A^2 = I$ by choosing nilpotent matrices $N$ such that $N^2 = 0$, and then setting $A = I + N$. If $N \neq 0$ (which is possible for $n\times n$ matrices with n > 1), then such an $A$ (identity plus non-zero nilpotent) is not diagonalizable (in any characteristic; but away from char. $2$, matrices of this form can't satisfy $A^2 = 0$).