[proofplan]
We first check that derivations form an $F$-vector subspace of $\operatorname{End}_F(\mathfrak h)$. Then we prove that the commutator of two derivations is again a derivation by expanding both compositions and cancelling the mixed terms. The Lie algebra identities for the commutator are verified directly inside the associative algebra $\operatorname{End}_F(\mathfrak h)$. Finally, the Jacobi identity in $\mathfrak h$ shows that each inner map $\operatorname{ad}_x$ satisfies the derivation rule.
[/proofplan]
[step:Show that derivations form an $F$-vector space]
Let $\operatorname{End}_F(\mathfrak h)$ denote the $F$-[vector space](/page/Vector%20Space) of all $F$-linear maps $\mathfrak h \to \mathfrak h$. The zero map $0_{\mathfrak h}:\mathfrak h \to \mathfrak h$ satisfies
\begin{align*}
0_{\mathfrak h}([u,v]_{\mathfrak h})
=
[0_{\mathfrak h}(u),v]_{\mathfrak h} + [u,0_{\mathfrak h}(v)]_{\mathfrak h}
\end{align*}
for all $u,v \in \mathfrak h$, so $0_{\mathfrak h}\in \operatorname{Der}(\mathfrak h)$.
Let $D_1,D_2 \in \operatorname{Der}(\mathfrak h)$ and let $a,b \in F$. Define the $F$-[linear map](/page/Linear%20Map)
\begin{align*}
D:\mathfrak h &\to \mathfrak h \\
u &\mapsto aD_1(u)+bD_2(u).
\end{align*}
For all $u,v \in \mathfrak h$, using the derivation identities for $D_1$ and $D_2$ and the bilinearity of $[\cdot,\cdot]_{\mathfrak h}$,
\begin{align*}
D([u,v]_{\mathfrak h})
&= aD_1([u,v]_{\mathfrak h}) + bD_2([u,v]_{\mathfrak h}) \\
&= a\bigl([D_1(u),v]_{\mathfrak h} + [u,D_1(v)]_{\mathfrak h}\bigr)
+ b\bigl([D_2(u),v]_{\mathfrak h} + [u,D_2(v)]_{\mathfrak h}\bigr) \\
&= [aD_1(u)+bD_2(u),v]_{\mathfrak h}
+ [u,aD_1(v)+bD_2(v)]_{\mathfrak h} \\
&= [D(u),v]_{\mathfrak h} + [u,D(v)]_{\mathfrak h}.
\end{align*}
Thus $D\in \operatorname{Der}(\mathfrak h)$, and $\operatorname{Der}(\mathfrak h)$ is an $F$-vector subspace of $\operatorname{End}_F(\mathfrak h)$.
[/step]
[step:Compute the derivation rule for the commutator]
Let $D_1,D_2 \in \operatorname{Der}(\mathfrak h)$. Define the commutator map
\begin{align*}
C:\mathfrak h &\to \mathfrak h \\
u &\mapsto D_1(D_2(u))-D_2(D_1(u)).
\end{align*}
Since $D_1$ and $D_2$ are $F$-linear, $C$ is $F$-linear.
Fix $u,v \in \mathfrak h$. First apply $D_2$ to $[u,v]_{\mathfrak h}$ and then apply $D_1$:
\begin{align*}
D_1(D_2([u,v]_{\mathfrak h}))
&= D_1\bigl([D_2(u),v]_{\mathfrak h} + [u,D_2(v)]_{\mathfrak h}\bigr) \\
&= [D_1(D_2(u)),v]_{\mathfrak h}
+ [D_2(u),D_1(v)]_{\mathfrak h} \\
&\quad + [D_1(u),D_2(v)]_{\mathfrak h}
+ [u,D_1(D_2(v))]_{\mathfrak h}.
\end{align*}
Similarly,
\begin{align*}
D_2(D_1([u,v]_{\mathfrak h}))
&= [D_2(D_1(u)),v]_{\mathfrak h}
+ [D_1(u),D_2(v)]_{\mathfrak h} \\
&\quad + [D_2(u),D_1(v)]_{\mathfrak h}
+ [u,D_2(D_1(v))]_{\mathfrak h}.
\end{align*}
Subtracting the second identity from the first cancels the two mixed terms and gives
\begin{align*}
C([u,v]_{\mathfrak h})
&= [C(u),v]_{\mathfrak h} + [u,C(v)]_{\mathfrak h}.
\end{align*}
Therefore $C\in \operatorname{Der}(\mathfrak h)$. Since $C=[D_1,D_2]_{\operatorname{Der}}$, the commutator bracket is closed on $\operatorname{Der}(\mathfrak h)$.
[guided]
The only point needing computation is closure under the commutator. We define the candidate commutator as the $F$-linear map
\begin{align*}
C:\mathfrak h &\to \mathfrak h \\
u &\mapsto D_1(D_2(u))-D_2(D_1(u)).
\end{align*}
To prove $C$ is a derivation, we must prove
\begin{align*}
C([u,v]_{\mathfrak h})
=
[C(u),v]_{\mathfrak h}+[u,C(v)]_{\mathfrak h}
\end{align*}
for arbitrary $u,v \in \mathfrak h$.
Start with the first composition. Since $D_2$ is a derivation,
\begin{align*}
D_2([u,v]_{\mathfrak h})
=
[D_2(u),v]_{\mathfrak h}+[u,D_2(v)]_{\mathfrak h}.
\end{align*}
Now apply $D_1$ to both bracket terms, using both $F$-linearity and the derivation rule for $D_1$:
\begin{align*}
D_1(D_2([u,v]_{\mathfrak h}))
&= D_1\bigl([D_2(u),v]_{\mathfrak h}+[u,D_2(v)]_{\mathfrak h}\bigr) \\
&= [D_1(D_2(u)),v]_{\mathfrak h}
+ [D_2(u),D_1(v)]_{\mathfrak h} \\
&\quad + [D_1(u),D_2(v)]_{\mathfrak h}
+ [u,D_1(D_2(v))]_{\mathfrak h}.
\end{align*}
Interchanging the roles of $D_1$ and $D_2$ gives
\begin{align*}
D_2(D_1([u,v]_{\mathfrak h}))
&= [D_2(D_1(u)),v]_{\mathfrak h}
+ [D_1(u),D_2(v)]_{\mathfrak h} \\
&\quad + [D_2(u),D_1(v)]_{\mathfrak h}
+ [u,D_2(D_1(v))]_{\mathfrak h}.
\end{align*}
The reason the commutator works is that the two mixed terms appear with the same sign in both expansions:
\begin{align*}
[D_2(u),D_1(v)]_{\mathfrak h}
\quad\text{and}\quad
[D_1(u),D_2(v)]_{\mathfrak h}.
\end{align*}
They therefore cancel after subtraction. Hence
\begin{align*}
C([u,v]_{\mathfrak h})
&= D_1(D_2([u,v]_{\mathfrak h}))-D_2(D_1([u,v]_{\mathfrak h})) \\
&= [D_1(D_2(u))-D_2(D_1(u)),v]_{\mathfrak h}
+ [u,D_1(D_2(v))-D_2(D_1(v))]_{\mathfrak h} \\
&= [C(u),v]_{\mathfrak h}+[u,C(v)]_{\mathfrak h}.
\end{align*}
Thus the commutator of two derivations is again a derivation.
[/guided]
[/step]
[step:Verify the Lie algebra identities for the commutator]
The bracket $[\cdot,\cdot]_{\operatorname{Der}}$ is bilinear because composition in $\operatorname{End}_F(\mathfrak h)$ is bilinear and subtraction is bilinear.
For $D \in \operatorname{Der}(\mathfrak h)$,
\begin{align*}
[D,D]_{\operatorname{Der}}
=
D\circ D-D\circ D
=
0_{\mathfrak h},
\end{align*}
so the bracket is alternating.
Let $D_1,D_2,D_3 \in \operatorname{Der}(\mathfrak h)$. Expanding the three commutators inside $\operatorname{End}_F(\mathfrak h)$ gives
\begin{align*}
[[D_1,D_2]_{\operatorname{Der}},D_3]_{\operatorname{Der}}
&= D_1D_2D_3-D_2D_1D_3-D_3D_1D_2+D_3D_2D_1, \\
[[D_2,D_3]_{\operatorname{Der}},D_1]_{\operatorname{Der}}
&= D_2D_3D_1-D_3D_2D_1-D_1D_2D_3+D_1D_3D_2, \\
[[D_3,D_1]_{\operatorname{Der}},D_2]_{\operatorname{Der}}
&= D_3D_1D_2-D_1D_3D_2-D_2D_3D_1+D_2D_1D_3.
\end{align*}
Here juxtaposition denotes composition of maps. Adding these identities, every triple composition occurs once with coefficient $1$ and once with coefficient $-1$. Hence
\begin{align*}
[[D_1,D_2]_{\operatorname{Der}},D_3]_{\operatorname{Der}}
+
[[D_2,D_3]_{\operatorname{Der}},D_1]_{\operatorname{Der}}
+
[[D_3,D_1]_{\operatorname{Der}},D_2]_{\operatorname{Der}}
=
0_{\mathfrak h}.
\end{align*}
Thus the Jacobi identity holds. Therefore $\operatorname{Der}(\mathfrak h)$ is a Lie algebra over $F$ under the commutator bracket.
[/step]
[step:Use the Jacobi identity to prove that inner adjoint maps are derivations]
Fix $x \in \mathfrak h$. Define
\begin{align*}
\operatorname{ad}_x:\mathfrak h &\to \mathfrak h \\
y &\mapsto [x,y]_{\mathfrak h}.
\end{align*}
The map $\operatorname{ad}_x$ is $F$-linear because the bracket $[\cdot,\cdot]_{\mathfrak h}$ is bilinear.
Let $y,z \in \mathfrak h$. The Jacobi identity in $\mathfrak h$ gives
\begin{align*}
[x,[y,z]_{\mathfrak h}]_{\mathfrak h}
+
[y,[z,x]_{\mathfrak h}]_{\mathfrak h}
+
[z,[x,y]_{\mathfrak h}]_{\mathfrak h}
=
0.
\end{align*}
Using skew-symmetry of the bracket,
\begin{align*}
[y,[z,x]_{\mathfrak h}]_{\mathfrak h}
=
-[y,[x,z]_{\mathfrak h}]_{\mathfrak h},
\qquad
[z,[x,y]_{\mathfrak h}]_{\mathfrak h}
=
-[[x,y]_{\mathfrak h},z]_{\mathfrak h}.
\end{align*}
Substituting these identities into Jacobi and rearranging yields
\begin{align*}
[x,[y,z]_{\mathfrak h}]_{\mathfrak h}
=
[[x,y]_{\mathfrak h},z]_{\mathfrak h}
+
[y,[x,z]_{\mathfrak h}]_{\mathfrak h}.
\end{align*}
In terms of $\operatorname{ad}_x$, this is
\begin{align*}
\operatorname{ad}_x([y,z]_{\mathfrak h})
=
[\operatorname{ad}_x(y),z]_{\mathfrak h}
+
[y,\operatorname{ad}_x(z)]_{\mathfrak h}.
\end{align*}
Since $y,z \in \mathfrak h$ were arbitrary, $\operatorname{ad}_x \in \operatorname{Der}(\mathfrak h)$. This proves the final assertion.
[/step]
