The extension basically is needed because not all polynomials factor completely over any field, for example x²+1 is not factorizable over the reals, but over the complex numbers you have x²+1 = (x-i)(x+i). Since the eigenvalues of any matrix are the roots of the characteristic polynomial, and in Jordan normal form the eigenvalues of your matrix are along the main diagonal, you need to include this.
To show that every matrix then has a Jordan normal form, you basically construct the normal form itself, which goes by induction (the section "A proof" in the wikipedia article gives a short overview).
However, for your original problem, you even don't need to construct the whole Jordan normal form. You only need to construct the Jordan block corresponding to λ=0, and leave the rest at it is (this rest gives you B). Then you don't even need to factor the characteristic polynomial corresponding to B, and so don't need to extend your field.