Derivations form a Lie algebra and inner derivations an ideal

Derivations form a linear subspace of ${\mathrm{End}}_k(\mathfrak{g})$, because the Leibniz identity is linear in $D$. For derivations $D,E$, expansion of $(DE-ED)[x,y]$ gives $[(DE-ED)x,y]+[x,(DE-ED)y]$: the two cross terms $[Dx,Ey]$ and $[Ex,Dy]$ occur once with each sign and cancel. Hence $[D,E]=DE-ED$ is a derivation.

Jacobi rewritten as $[x,[y,z]]-[y,[x,z]]=[[x,y],z]$ says $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$. It also says ${\mathrm{ad}}_x[y,z]=[{\mathrm{ad}}_{x}y,z]+[y,{\mathrm{ad}}_{x}z]$, so every ${\mathrm{ad}}_x$ is a derivation and $\mathrm{ad}$ is a Lie homomorphism.

The endomorphism commutator is bilinear and alternating, and its Jacobi identity follows by expanding the six triple composites. Therefore the closed linear subspace in step 1.1 is a Lie subalgebra.

If $D$ is any derivation, then for every $y$, $[D,{\mathrm{ad}}_x](y)=D[x,y]-[x,Dy]=[Dx,y]$; hence $[D,{\mathrm{ad}}_x]={\mathrm{ad}}_{Dx}$. Thus the inner derivations form an ideal of $\mathrm{Der}(\mathfrak{g})$.

Finally, $x\in \ker (\mathrm{ad})$ exactly when $[x,y]=0$ for every $y$, which is exactly $x\in Z(\mathfrak{g})$ by [L3]. For an abelian algebra the inner ideal is zero; for the zero algebra all assertions remain valid.