= Solution
A finite-dimensional <Lie algebra> $L$ is <Semisimple Lie algebra>[semisimple] when its <solvable radical> is zero, equivalently when it has no nonzero solvable <ideal of a Lie algebra>[ideals].
We prove the <Weyl complete reducibility theorem>. Induct on the dimension of a finite-dimensional $L$-module $V$. It is enough first to split a submodule $W$ of codimension one. The one-dimensional quotient is trivial because a semisimple Lie algebra is <perfect Lie algebra>[perfect]. By induction, $W$ is a direct sum of irreducible modules. A <Casimir element> formed using the <Killing form> commutes with the $L$-action, acts as zero on every trivial summand, and acts by a nonzero scalar on every nontrivial irreducible summand. Its image is therefore the sum $W_1$ of the nontrivial summands, while its kernel contains the trivial summands $W_0$ and maps onto $V/W$. Thus
$$
V=W_1\oplus\ker\Omega.
$$
Inside $\ker\Omega$, choose a lift $v$ of a basis of $V/W$. For $x\in L$, $xv\in W_0$, and $L$ acts trivially on $W_0$. Hence $[x,y]v=0$ for all $x,y\in L$. Since $L=[L,L]$, actually $xv=0$ for every $x$, so $\mathbb Cv$ is the required invariant complement.
This codimension-one case implies the general case. For an arbitrary submodule $W\subseteq V$, let
$$
X=\{f\in\operatorname{Hom}_{\mathbb C}(V,W):f|_W\text{ is scalar}\}
$$
with the natural <Hom representation>. The maps vanishing on $W$ form an $L$-submodule $X_0$ of codimension one. Splitting $X_0$ supplies an $L$-equivariant $f$ with $f|_W=I_W$. Then
$$
V=W\oplus\ker f,
$$
so every invariant subspace has an invariant complement and every finite-dimensional representation is completely reducible.
For the requested example, embed $\mathfrak{sl}_2$ as the upper-left $2\times2$ block in the <Special linear Lie algebra> $\mathfrak{sl}_3$. Under the restricted <Adjoint representation>,
$$
\mathfrak{sl}_3
=\underbrace{\mathfrak{sl}_2}_{V_2}
\oplus\underbrace{\mathbb C\operatorname{diag}(1,1,-2)}_{V_0}
\oplus\underbrace{\langle E_{13},E_{23}\rangle}_{V_1}
\oplus\underbrace{\langle E_{31},E_{32}\rangle}_{V_1}.
$$
Here $V_n$ is the irreducible $\mathfrak{sl}_2$-module of highest weight $n$. The first summand is the three-dimensional adjoint module, the second is trivial, and the last two are the two-dimensional defining module and its dual, which are isomorphic. This explicit direct sum demonstrates complete reducibility.
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