OK, let me try again. First, a general construction. Let V be an object in an abelian category, A = End(V) and J a finitely generated two-sided ideal of A, with generators (j_1, ..., j_r). Define V/JV to be the cokernel of V^{r} --> V, where the map is multiplication by j_i on the ith coordinate. We need to check that this depends only on J and not on the choice of generators; I haven't actually done this. Define JV to be the kernel of V --> V/JV. So we have a short exact sequence

0 --> JV --> V --> V/JV --> 0

I claim that the action of A on V passes to an action of A/J on V/JV. Also, A acts on JV. (One can say more than this, but we don't need to.)

Using the action of A on JV, we can repeat this construction to get JV/J^2V, J^2 V/J^3 V, etcetera. All of these come with actions of A/J.

Now, suppose that all of our Hom spaces are finite dimensional. A finite dimensional algebra is semi-simple if and only if it has no nontrivial nilpotent two-sided ideal. So, suppose for the sake of contradiction that there is some (V,A,J) as above with J nilpotent. Then J^k V is eventually zero. Trace is additive in short exact sequences, so

Tr(f: V --> V) = \sum Tr(f: J^k V/J^{k+1} V --> J^k V/J^{k+1} V).

If f is in J, the right hand side is 0. Also, if f is in J, so is fg for any g in A because J is an ideal. So J is in the kernel of the trace pairing, and we deduce that J=0.

So all endomorphism rings are semi-simple and, by the lemma cited above, so is the category.