A crucial step in the "purely algebraic" proof of Weyl's semisimplicity theorem is that the Casimir element $C\in U\mathfrak{g}$ acts by nonzero scalars on a nontrivial irrep $V$. However, at least two sources I have consulted assert that $\text{tr}_V(C)=\text{tr}(C)=\dim \mathfrak{g}$, i.e. using the fact that $C$ can be written as a sum of products dual basis elements and then asserting that the trace pairing on the irrep is the same as the Killing form. But this seems obviously false. Or at least, I don't see why it should be true.

What I can see is that if $\mathfrak{g}$ is simple, the trace pairing on the irrep is a scaling of the Killing form, so by picking a suitable basis, we get that the trace pairing on $V$ is a scaling of the sum of the Killing forms on the simple summands of semisimple $\mathfrak{g}$, so you get a linear combination of the dimensions of the simple summands. But this could very well be zero.