In addition to the other answers, all of which are quite good, I offer a rather pedestrian observation: If you perturb the diagonal in each Jordan block of your given matrix $T$ so all the diagonal terms have different values, you end up with a matrix that has $n$ distinct eigenvalues and is hence diagonalizable. Such a perturbation can of course be as small as you wish.

Edit: As gowers points out, you don't even need the Jordan form to do this, just the triangular form.