First Whitehead lemma

On $E=M\oplus k$ define $x\cdot (m,a)=(xm+a\delta (x),0)$. The commutator of the actions at $(m,a)$ has first component $[x,y]m+a(x\delta (y)-y\delta (x))$, which equals $[x,y]m+a\delta ([x,y])$ by [L1]. Thus this is a representation, $M$ is a submodule, and $E\mathrm{/}M$ is the trivial one-dimensional module.

By [L2], $M$ has an invariant line complement $L$. Its projection to $k$ is an isomorphism, so $L$ has a generator $(m,1)$. A one-dimensional module kills the derived algebra, which is all of $\mathfrak{g}$ by [L3]; hence it is trivial and $0=x\cdot (m,1)=(xm+\delta (x),0)$. Therefore $\delta (x)=x(-m)$ is a coboundary by [L1].

Every cocycle is therefore a coboundary and the quotient $H^1$ is zero. If $M=0$ or $\mathfrak{g}=0$, the same conclusion is immediate (the zero algebra is semisimple and has no nonzero $1$-cochains into a zero module; for $\mathfrak{g}=0$ the Hom space itself is zero). The complement in step 2.1 is supplied by the proved finite-dimensional theorem, not by a choice principle.