This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem.
General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is a faithful irreducible rep and $k$ has a trivial action. The argument goes that the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a section as the kernel of the Casimir element.
The reason we should expect this to be so is that the Casimir element is really the Laplacian of the Lie group with bi-invariant metric coming from the trace form. By the Peter-Weyl theorem, every irreducible representation of a compact Lie group $G$ embeds into $L^2(G)$ with the translation action.
On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constant ones since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, it must act by a nonzero scalar, hence the argument for Weyl's theorem should work.