[proofplan]
We first verify that derivations form a vector subspace of $\operatorname{End}_F(\mathfrak h)$ because the Leibniz identity is linear in the operator. We then prove closure under the commutator by expanding $(D_1 \circ D_2 - D_2 \circ D_1)([u,v]_{\mathfrak h})$ and cancelling the mixed terms. The remaining Lie algebra axioms for the commutator follow from bilinearity and associativity of composition. Finally, the Jacobi identity in $\mathfrak h$ gives exactly the Leibniz rule for each adjoint map $\operatorname{ad}_x$.
[/proofplan]
[step:Show that derivations form a vector subspace of $\operatorname{End}_F(\mathfrak h)$]
Let $0_{\operatorname{End}}: \mathfrak h \to \mathfrak h$ denote the zero $F$-[linear map](/page/Linear%20Map). For all $u,v \in \mathfrak h$,
\begin{align*}
0_{\operatorname{End}}([u,v]_{\mathfrak h})
=
0_{\mathfrak h}
=
[0_{\operatorname{End}}(u),v]_{\mathfrak h}
+
[u,0_{\operatorname{End}}(v)]_{\mathfrak h},
\end{align*}
so $0_{\operatorname{End}} \in \operatorname{Der}(\mathfrak h)$.
Let $D,E: \mathfrak h \to \mathfrak h$ be elements of $\operatorname{Der}(\mathfrak h)$, and let $a,b \in F$. Define the $F$-linear map $aD+bE: \mathfrak h \to \mathfrak h$ by
\begin{align*}
(aD+bE)(w) := aD(w)+bE(w)
\end{align*}
for $w \in \mathfrak h$. For all $u,v \in \mathfrak h$, using the derivation identities for $D$ and $E$ and the bilinearity of $[\cdot,\cdot]_{\mathfrak h}$,
\begin{align*}
(aD+bE)([u,v]_{\mathfrak h})
&=
aD([u,v]_{\mathfrak h})+bE([u,v]_{\mathfrak h})\\
&=
a\bigl([D(u),v]_{\mathfrak h}+[u,D(v)]_{\mathfrak h}\bigr)
+
b\bigl([E(u),v]_{\mathfrak h}+[u,E(v)]_{\mathfrak h}\bigr)\\
&=
[aD(u)+bE(u),v]_{\mathfrak h}
+
[u,aD(v)+bE(v)]_{\mathfrak h}\\
&=
[(aD+bE)(u),v]_{\mathfrak h}
+
[u,(aD+bE)(v)]_{\mathfrak h}.
\end{align*}
Thus $aD+bE \in \operatorname{Der}(\mathfrak h)$, and $\operatorname{Der}(\mathfrak h)$ is an $F$-vector subspace of $\operatorname{End}_F(\mathfrak h)$.
[/step]
[step:Prove that the commutator of two derivations is a derivation]
Let $D,E: \mathfrak h \to \mathfrak h$ be elements of $\operatorname{Der}(\mathfrak h)$. Define the commutator map $[D,E]_{\operatorname{End}}: \mathfrak h \to \mathfrak h$ by
\begin{align*}
[D,E]_{\operatorname{End}} := D \circ E - E \circ D.
\end{align*}
This map is $F$-linear because $D$, $E$, and composition of $F$-linear maps are $F$-linear.
Fix $u,v \in \mathfrak h$. Applying the derivation rule for $E$ and then for $D$ gives
\begin{align*}
D(E([u,v]_{\mathfrak h}))
&=
D\bigl([E(u),v]_{\mathfrak h}+[u,E(v)]_{\mathfrak h}\bigr)\\
&=
[D(E(u)),v]_{\mathfrak h}
+
[E(u),D(v)]_{\mathfrak h}
+
[D(u),E(v)]_{\mathfrak h}
+
[u,D(E(v))]_{\mathfrak h}.
\end{align*}
Similarly,
\begin{align*}
E(D([u,v]_{\mathfrak h}))
&=
E\bigl([D(u),v]_{\mathfrak h}+[u,D(v)]_{\mathfrak h}\bigr)\\
&=
[E(D(u)),v]_{\mathfrak h}
+
[D(u),E(v)]_{\mathfrak h}
+
[E(u),D(v)]_{\mathfrak h}
+
[u,E(D(v))]_{\mathfrak h}.
\end{align*}
Subtracting the second identity from the first, the two mixed terms cancel:
\begin{align*}
[D,E]_{\operatorname{End}}([u,v]_{\mathfrak h})
&=
[D(E(u))-E(D(u)),v]_{\mathfrak h}
+
[u,D(E(v))-E(D(v))]_{\mathfrak h}\\
&=
[[D,E]_{\operatorname{End}}(u),v]_{\mathfrak h}
+
[u,[D,E]_{\operatorname{End}}(v)]_{\mathfrak h}.
\end{align*}
Therefore $[D,E]_{\operatorname{End}} \in \operatorname{Der}(\mathfrak h)$.
[guided]
We must check the defining Leibniz rule for the new map $D \circ E - E \circ D$. Let $D,E: \mathfrak h \to \mathfrak h$ be derivations, and define
\begin{align*}
[D,E]_{\operatorname{End}}: \mathfrak h &\to \mathfrak h,\\
w &\mapsto D(E(w))-E(D(w)).
\end{align*}
The map is $F$-linear because it is a difference of compositions of $F$-linear maps.
Now fix $u,v \in \mathfrak h$. The only point is to expand carefully enough that the cancellation is visible. Since $E$ is a derivation,
\begin{align*}
E([u,v]_{\mathfrak h})
=
[E(u),v]_{\mathfrak h}
+
[u,E(v)]_{\mathfrak h}.
\end{align*}
Applying $D$ to this identity and using that $D$ is a derivation on each bracket term,
\begin{align*}
D(E([u,v]_{\mathfrak h}))
&=
D\bigl([E(u),v]_{\mathfrak h}+[u,E(v)]_{\mathfrak h}\bigr)\\
&=
[D(E(u)),v]_{\mathfrak h}
+
[E(u),D(v)]_{\mathfrak h}
+
[D(u),E(v)]_{\mathfrak h}
+
[u,D(E(v))]_{\mathfrak h}.
\end{align*}
Interchanging the roles of $D$ and $E$ gives
\begin{align*}
E(D([u,v]_{\mathfrak h}))
&=
E\bigl([D(u),v]_{\mathfrak h}+[u,D(v)]_{\mathfrak h}\bigr)\\
&=
[E(D(u)),v]_{\mathfrak h}
+
[D(u),E(v)]_{\mathfrak h}
+
[E(u),D(v)]_{\mathfrak h}
+
[u,E(D(v))]_{\mathfrak h}.
\end{align*}
Subtracting is exactly where the commutator matters. The mixed terms $[E(u),D(v)]_{\mathfrak h}$ and $[D(u),E(v)]_{\mathfrak h}$ occur once in each expansion with opposite signs, so they cancel. Hence
\begin{align*}
[D,E]_{\operatorname{End}}([u,v]_{\mathfrak h})
&=
D(E([u,v]_{\mathfrak h}))-E(D([u,v]_{\mathfrak h}))\\
&=
[D(E(u))-E(D(u)),v]_{\mathfrak h}
+
[u,D(E(v))-E(D(v))]_{\mathfrak h}\\
&=
[[D,E]_{\operatorname{End}}(u),v]_{\mathfrak h}
+
[u,[D,E]_{\operatorname{End}}(v)]_{\mathfrak h}.
\end{align*}
This is precisely the derivation identity for $[D,E]_{\operatorname{End}}$, so the commutator of two derivations is again a derivation.
[/guided]
[/step]
[step:Verify the Lie algebra identities for the commutator bracket]
Define
\begin{align*}
[\cdot,\cdot]_{\operatorname{Der}}:
\operatorname{Der}(\mathfrak h) \times \operatorname{Der}(\mathfrak h)
&\to \operatorname{Der}(\mathfrak h),\\
(D,E)
&\mapsto D \circ E - E \circ D.
\end{align*}
The previous step proves that this map has codomain $\operatorname{Der}(\mathfrak h)$.
Bilinearity follows from the bilinearity of addition, scalar multiplication, and composition in $\operatorname{End}_F(\mathfrak h)$. For every $D \in \operatorname{Der}(\mathfrak h)$,
\begin{align*}
[D,D]_{\operatorname{Der}}
=
D \circ D - D \circ D
=
0_{\operatorname{End}},
\end{align*}
so the bracket is alternating.
It remains to verify the Jacobi identity. Let $D,E,G \in \operatorname{Der}(\mathfrak h)$. Using associativity of composition in $\operatorname{End}_F(\mathfrak h)$, we expand:
\begin{align*}
[D,[E,G]_{\operatorname{Der}}]_{\operatorname{Der}}
&=
D \circ E \circ G
-
D \circ G \circ E
-
E \circ G \circ D
+
G \circ E \circ D,\\
[E,[G,D]_{\operatorname{Der}}]_{\operatorname{Der}}
&=
E \circ G \circ D
-
E \circ D \circ G
-
G \circ D \circ E
+
D \circ G \circ E,\\
[G,[D,E]_{\operatorname{Der}}]_{\operatorname{Der}}
&=
G \circ D \circ E
-
G \circ E \circ D
-
D \circ E \circ G
+
E \circ D \circ G.
\end{align*}
Adding the three displayed identities cancels every composition term with the same term of opposite sign. Therefore
\begin{align*}
[D,[E,G]_{\operatorname{Der}}]_{\operatorname{Der}}
+
[E,[G,D]_{\operatorname{Der}}]_{\operatorname{Der}}
+
[G,[D,E]_{\operatorname{Der}}]_{\operatorname{Der}}
=
0_{\operatorname{End}}.
\end{align*}
Thus $\operatorname{Der}(\mathfrak h)$ is a Lie algebra over $F$ under the commutator bracket.
[/step]
[step:Use the Jacobi identity to prove that every adjoint map is a derivation]
Fix $x \in \mathfrak h$. Define the adjoint map
\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_{\mathfrak h}.
\end{align*}
Since the bracket is alternating and bilinear, it is skew-symmetric, so
\begin{align*}
[y,[z,x]_{\mathfrak h}]_{\mathfrak h}
&=
-[y,[x,z]_{\mathfrak h}]_{\mathfrak h},\\
[z,[x,y]_{\mathfrak h}]_{\mathfrak h}
&=
-[[x,y]_{\mathfrak h},z]_{\mathfrak h}.
\end{align*}
Substituting these identities into Jacobi and rearranging gives
\begin{align*}
\operatorname{ad}_x([y,z]_{\mathfrak h})
&=
[x,[y,z]_{\mathfrak h}]_{\mathfrak h}\\
&=
[[x,y]_{\mathfrak h},z]_{\mathfrak h}
+
[y,[x,z]_{\mathfrak h}]_{\mathfrak h}\\
&=
[\operatorname{ad}_x(y),z]_{\mathfrak h}
+
[y,\operatorname{ad}_x(z)]_{\mathfrak h}.
\end{align*}
This is the derivation identity for $\operatorname{ad}_x$, hence $\operatorname{ad}_x \in \operatorname{Der}(\mathfrak h)$. The two assertions of the theorem follow.
[/step]
