= Solution
Use the finite-dimensional complex-module setting of the course. We prove the <Weyl complete reducibility theorem> by <Haar averaging produces invariant complements>, defining its constructions rather than invoking complete reducibility itself.
Choose a <compact real form> $\mathfrak k\subset\mathfrak g$: it is a real Lie subalgebra with $\mathfrak g=\mathfrak k\oplus i\mathfrak k$, and its simply connected Lie group $K$ is compact. Existence of this form and compact group is a standard fact for complex <semisimple Lie algebras>. Restrict the representation to $\mathfrak k$; <integration of a Lie-algebra representation> gives a group representation $R:K\to GL(V)$. Since $K$ is connected and generated by exponentials, the $\mathfrak g$-invariant subspace $W$ is also $K$-invariant.
Choose any positive-definite <Hermitian inner product> on $V$. With the <Haar measure> of $K$ normalized to mass one, define
$$
\langle v,w\rangle_K=\int_K\langle R(k)v,R(k)w\rangle_0\,dk.
$$
This is a Hermitian inner product, and it is positive-definite: for $v\ne0$ every $R(k)v$ is nonzero, so the continuous positive integrand has positive integral. Right invariance of Haar measure gives $\langle R(a)v,R(a)w\rangle_K=\langle v,w\rangle_K$ for $a\in K$.
Let $W'=W^\perp$ for this averaged inner product. Finite-dimensional inner-product linear algebra gives $V=W\oplus W'$. If $v\in W'$ and $w\in W$, then
$$
\langle R(a)v,w\rangle_K=\langle v,R(a^{-1})w\rangle_K=0,
$$
because $R(a^{-1})w\in W$. Thus $W'$ is $K$-invariant. Differentiating makes it $\mathfrak k$-invariant, and since it is a complex vector subspace, $\mathfrak g=\mathfrak k+i\mathfrak k$ also preserves it. Therefore
$$
\boxed{W'=W^{\perp_K}\text{ is a }\mathfrak g\text{-submodule and }V=W\oplus W'.}
$$
The zero or whole-space submodule cases are included. This proves the required complement theorem without circularly assuming it.
The finite-dimensional hypothesis is essential to the statement. The <nonsplit zero-weight Verma extension for sl2> is a counterexample for unrestricted modules: $M(0)$ has basis $v_n$ with $hv_n=-2nv_n$, $fv_n=v_{n+1}$ and $ev_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge1$ has trivial one-dimensional quotient. A complement would be trivial and contained in the zero-weight line $\mathbb Cv_0$, yet $fv_0=v_1\ne0$. Hence no such complement exists in that infinite-dimensional case. The paper's finite-dimensional representation setting supplies the necessary hypothesis.
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