= Solution
The defining rule for a <derivation of a Lie algebra> is linear in the operator, so $\operatorname{Der}(\mathfrak g)$ is a <vector subspace> of the <general linear Lie algebra> $\mathfrak{gl}(\mathfrak g)$. Its <Lie bracket> is the operator <commutator>. If $D,E$ are <derivations of a Lie algebra>, expand both compositions:
$$
\begin{aligned}
DE[x,y]&=[DEx,y]+[Ex,Dy]+[Dx,Ey]+[x,DEy],\\
ED[x,y]&=[EDx,y]+[Dx,Ey]+[Ex,Dy]+[x,EDy].
\end{aligned}
$$
Subtracting cancels the mixed terms and gives
$$
[D,E][x,y]=[[D,E]x,y]+[x,[D,E]y].
$$
Thus $[D,E]$ is again a <derivation of a Lie algebra>, proving that the <derivation Lie algebra> is a <Lie subalgebra> of $\mathfrak{gl}(\mathfrak g)$. This calculation works in every <characteristic of a field>.
The <Jacobi identity> states $[x,[y,z]]=[[x,y],z]+[y,[x,z]]$. It first shows that $\operatorname{ad}_x$ is a <derivation of a Lie algebra>, and then gives
$$
[\operatorname{ad}_x,\operatorname{ad}_y](z)=[x,[y,z]]-[y,[x,z]]=[[x,y],z].
$$
The map $x\mapsto\operatorname{ad}_x$ is linear, so this identity proves it is a <Lie algebra homomorphism>.
Finally, for any <derivation of a Lie algebra> $D$,
$$
[D,\operatorname{ad}_x](y)=D[x,y]-[x,Dy]=[Dx,y].
$$
Consequently
$$
\boxed{[D,\operatorname{ad}_x]=\operatorname{ad}_{Dx},\qquad[\operatorname{Der}(\mathfrak g),\operatorname{ad}\mathfrak g]\subseteq\operatorname{ad}\mathfrak g.}
$$
This proves that the <inner derivations of a Lie algebra> form an <ideal of a Lie algebra> in $\operatorname{Der}(\mathfrak g)$.
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