= Solution
A <Lie algebra representation> $V$ is completely reducible when it is a <direct sum> of irreducible <submodules>. In finite dimension this is equivalent to saying that every <submodule> $W\subseteq V$ has an invariant complementary subspace. Indeed, one may split off a minimal nonzero <submodule> and induct on dimension; conversely, if $V$ is a finite sum of simple <submodules>, choose a maximal sum $U$ of these with $U\cap W=0$. If $W+U\ne V$, some simple summand $S$ is not contained in $W+U$. Then $S\cap(W+U)=0$ by simplicity, so $U+S$ is a larger sum disjoint from $W$, a contradiction. Thus $V=W\oplus U$.
We prove the required splitting by <Haar averaging produces invariant complements>. The structural input is the <compact real form of a complex semisimple Lie algebra>: there is a real semisimple algebra $\mathfrak k$ with $L=\mathfrak k\oplus i\mathfrak k$, whose simply connected integrating Lie group $K$ is compact. This structural fact is independent of the complete-reducibility assertion. One standard construction uses <root vectors> normalized so that $[e_\alpha,e_{-\alpha}]=h_\alpha$, with real structure constants compatible with the involution $e_\alpha\mapsto-e_{-\alpha}$ and $h_\alpha\mapsto-h_\alpha$. The real span
$$
\mathfrak k=\operatorname{span}_{\mathbb R}\{ih_i,\ e_\alpha-e_{-\alpha},\ i(e_\alpha+e_{-\alpha}):\alpha>0\}
$$
is closed under brackets, complexifies to $L$, and has negative-definite <Killing form>. The compactness criterion for real semisimple <Lie algebras> then gives a compact simply connected $K$. Thus this route uses root structure and compact integration, not the theorem being proved.
Restrict $\rho:L\to\operatorname{End}_{\mathbb C}(V)$ to $\mathfrak k$. The <integration of a Lie-algebra representation> gives a representation $R:K\to\operatorname{GL}_{\mathbb C}(V)$, since $K$ is simply connected. Choose any positive-definite <Hermitian inner product> $\langle\ ,\ \rangle_0$ and average using normalized <Haar measure>:
$$
\langle v,w\rangle_K=\int_K\langle R(g)v,R(g)w\rangle_0\,d\mu(g).
$$
The integral exists by compactness. It is positive-definite because $R(g)v\ne0$ for $v\ne0$, and it is $K$-invariant by translation invariance of <Haar measure>. Equivalently, every $\rho(a)$ with $a\in\mathfrak k$ is skew-Hermitian for this inner product.
Let $W$ be an $L$-<submodule>. It is invariant under $\mathfrak k$, hence under the exponentials generating the connected group $K$. For $v\in W^\perp$, $w\in W$ and $g\in K$,
$$
\langle R(g)v,w\rangle_K=\langle v,R(g^{-1})w\rangle_K=0.
$$
Consequently the <orthogonal complement> $W^\perp$ is $K$-invariant, and differentiation makes it $\mathfrak k$-invariant. It is a complex vector subspace, so it is also invariant under $i\mathfrak k$ and hence under $L$. Therefore $V=W\oplus W^\perp$ as $L$-<modules>.
Induction on $\dim V$ now expresses $V$ as a finite <direct sum> of irreducible <submodules>. \b[Every finite-dimensional representation of a finite-dimensional complex semisimple <Lie algebra> is completely reducible.] The compact real form need not be the real form originally used to present $L$; in particular the averaging argument does not require the given presentation to be unitary.
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