= Solution
The one-dimensional quotient $V/W$ is trivial because every one-dimensional representation vanishes on the <derived series of a Lie algebra>[derived algebra] $[\mathfrak{sl}_2,\mathfrak{sl}_2]=\mathfrak{sl}_2$. Choose $v\in V$ mapping to $1\in V/W$. Then
$$
c(x)=xv\in W
$$
is a $1$-cocycle:
$$
c([x,y])=x,c(y)-y,c(x).
$$
We show that it is a coboundary.
Decompose $W$ into generalized eigenspaces of its <Casimir element> $\Omega$. These are subrepresentations because $\Omega$ is central. On a generalized eigenspace with nonzero eigenvalue, $\Omega$ is invertible. If $(x_i)$ and $(x^i)$ are dual bases of $\mathfrak{sl}_2$ for the <Killing form>, put
$$
u=\sum_i x_i c(x^i).
$$
Invariance of the Killing form and the cocycle identity give the standard Casimir calculation
$$
x u=\Omega c(x).
$$
Thus on every nonzero generalized eigenspace, $c(x)=x(\Omega^{-1}u)$.
On the zero generalized eigenspace, every irreducible composition factor has zero Casimir eigenvalue. By part i and the <Classification of finite-dimensional sl2 representations>, each such factor is trivial. In a basis adapted to a composition series, the image of $\mathfrak{sl}_2$ is therefore strictly upper triangular and hence solvable. Since $\mathfrak{sl}_2$ is simple and non-solvable, that image is zero. The cocycle then vanishes because it kills $[\mathfrak{sl}_2,\mathfrak{sl}_2]$.
Combining the generalized eigenspaces gives $w\in W$ such that $c(x)=xw$ for every $x$. Hence $v-w$ is invariant, and
$$
V=W\oplus\mathbb C(v-w)
$$
is a decomposition into subrepresentations. This proves the codimension-one case of the <Weyl complete reducibility theorem>.
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