Weyl's theorem states that any finite-dimensional representation of a finite-dimensional semisimple Lie algebra is completely reducible.  In my mind, the "natural" way to prove this result is by way of Lie groups.  However, as a student, I first encountered Weyl's theorem in the textbook by Humphreys, in which he gives a purely algebraic proof.  I remember finding this proof very mysterious, and in particular it seemed that Humphreys pulled the Casimir element out of a hat.  Looking at the proof again now, I still find it somewhat mysterious, and if I had to present the proof in a graduate class, I don't think I would be able to motivate it very well.

How would you motivate this purely algebraic proof of Weyl's theorem?

I think that there are two things to be motivated: one is the Casimir, and the other is the proof of semi-simplicity.

First, for the Casimir, it might help to note that there is a ``formula-free" construction. A symmetric bilinear form $\kappa$ defines $\mathfrak{g}\simeq \mathfrak{g}^{\vee}$, and the Casimir is the image in $U(\mathfrak{g})$ of the element corresponding to the identity under the isomorphism $\mathfrak{g}\otimes\mathfrak{g}\simeq \mathfrak{g}^{\vee}\otimes\mathfrak{g}$.

The idea of the proof is actually very simple: you construct an element of the center of the algebra which ``detects" the trivial representation. I.e., it acts by zero on the trivial representation and by a non-zero scalar on all simple (finite-dimensional) representations. (This is in exact analogy to what happens for finite groups in characteristic zero: then $1-\frac{1}{|G|}\sum_{g\in{G}}\delta_g$ has the same property).

Once you have such a central element, let's call it $C$, the proof that finite-dimensional representations are semi-simple is easy. For all $V,W$, note that:
$$\operatorname{Ext}^i(V,W)=
\operatorname{Ext}^i(\mathbb{C},\underline{\operatorname{Hom}}(V,W))$$
(where $\underline{\operatorname{Hom}}$ is internal $\operatorname{Hom}$ relative to the usual tensor product of $\mathfrak{g}$-modules, and both $\operatorname{Ext}$s are of $\mathfrak{g}$-modules) because formation of internal $Hom$ is exact in both variables. Therefore, it's enough to show that $\operatorname{Ext}^1(\mathbb{C},V)=0$ for all finite-dimensional $\mathfrak{g}$-modules $V$ (substitute $\underline{\operatorname{Hom}}(V,W)$ for $V$).

Clearly any such $V$ has finite length, so by devissage, it's enough to prove for simple modules. Either $V$ is trivial or it is not. If $V$ is non-trivial, then $C$ acts on $\operatorname{Ext}^i(\mathbb{C},V)$ by two different scalars: the one by which it acts on $V$ and by $0$ (by which it acts on $\mathbb{C}$). Therefore, this vector space must be zero. If $V=\mathbb{C}$, then $\operatorname{Ext}^1(\mathbb{C},\mathbb{C})=0$ since for any extension $E$, the homomorphism $\mathfrak{g}\to\operatorname{End}(E)$ maps to the 1-dimensional subspace sending $E$ to $\mathbb{C}$ and sending $\mathbb{C}$ to $0$, but since $\mathfrak{g}$ has no codimension 1 ideals this must be the trivial homomorphism.

Of course, basically the same proof goes through for finite groups and is essentially the same as the usual proof.
