= Solution
\b[Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb C$, every endomorphism of a finite-dimensional irreducible representation is scalar.] To prove the <Schur lemma>, let $T:V\to W$ be a <intertwining operator> between <irreducible representations>. Its <kernel> and <image of a linear map> are invariant. If $T\ne0$, irreducibility forces $\ker T=0$ and $\operatorname{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an <irreducible representation> form a <division ring>.
When $V$ is finite-dimensional over an <algebraically closed field>, an endomorphism $T$ has an <eigenvalue> $\lambda$. The endomorphism $T-\lambda I$ has nonzero <kernel>. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional <irreducible representations>, the space of <intertwining operators> has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.
\b[Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible.] We prove the <Weyl complete reducibility theorem> using the allowed <Casimir operator> facts, without assuming a splitting in advance. For dual <bases> $x_i,x^i$ with respect to the <Killing form>, the <Casimir element>
$$
\Omega=\sum_i x_i x^i\in U(\mathfrak g)
$$
is central in the <universal enveloping algebra>. Hence its <Casimir operator> $C_M=\sum_i\rho_M(x_i)\rho_M(x^i)$ commutes with the action on every <Lie algebra representation> $M$ and is compatible with <subrepresentations>, <quotient representations>, and <intertwining operators>. On the <trivial Lie algebra representation> it acts by zero. On every nontrivial finite-dimensional <Irreducible Lie algebra representation> it acts by a nonzero scalar.
For clarity, the last fact can be expressed by the <Casimir eigenvalue> formula: on an irreducible with <dominant integral weight> $\lambda$, the scalar is $(\lambda,\lambda+2\rho)$, with the inner product induced by the <Killing form> and $\rho$ the <half-sum of positive roots>. On the real span of the <weights> this inner product is positive definite, and the scalar is positive for $\lambda\ne0$. For a <semisimple Lie algebra> with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the <Casimir operator> being used here.
First establish <Casimir splitting of a trivial quotient>. Suppose
$$
0\longrightarrow N\longrightarrow E\overset{q}{\longrightarrow}\mathbb C\longrightarrow0
$$
is a <short exact sequence> of finite-dimensional <Lie algebra representations>, with trivial quotient. The <generalized eigenspaces> of $C_E$ are invariant, so
$$
E=E_0\oplus\bigoplus_{c\ne0}E_c.
$$
Each $E_c$ with $c\ne0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb C}=0$ gives $(-c)^r q(e)=0$. Thus $q(E_0)=\mathbb C$.
Take a <composition series of a module> of $E_0$. The <Casimir operator> is nilpotent on $E_0$, so its scalar on every irreducible <composition factor> is zero. The stated <Casimir operator> property makes every such factor trivial. In a <basis> adapted to the <composition series of a module>, the image of $\mathfrak g$ on $E_0$ therefore consists of strictly upper triangular <matrices>, so that image is <solvable Lie algebra>[solvable]. The allowed fact that a <semisimple Lie algebra> acts trivially on every one-dimensional representation implies that $\mathfrak g$ is a <perfect Lie algebra>: otherwise a nonzero <linear functional> on $\mathfrak g/[\mathfrak g,\mathfrak g]$ would define a nontrivial one-dimensional <Lie algebra representation>. Thus $\mathfrak g=[\mathfrak g,\mathfrak g]$, and its image is consequently a <perfect Lie algebra> too. A <perfect Lie algebra> that is also a <solvable Lie algebra> is zero, since its <derived series of a Lie algebra> is constant until it vanishes. Hence $\mathfrak g$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the <split short exact sequence> assertion.
Now let $W\subseteq V$ be any nonzero <subrepresentation>. On the <Hom representation> $\operatorname{Hom}(V,W)$ the action is
$$
(x\cdot f)(v)=x f(v)-f(xv).
$$
Consider the invariant subspace
$$
E=\{f\in\operatorname{Hom}(V,W):f|_W\in\mathbb C I_W\}.
$$
Restriction produces a <short exact sequence>
$$
0\longrightarrow\operatorname{Hom}(V/W,W)\longrightarrow E\longrightarrow\mathbb C I_W\longrightarrow0.
$$
Surjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding <Casimir splitting of a trivial quotient> yields an invariant $p:V\to W$ with $p|_W=I_W$. Thus
$$
V=W\oplus\ker p,
$$
and $\ker p$ is invariant. The zero <subrepresentation> also has a complement. Repeatedly splitting off an irreducible <subrepresentation> now gives a <direct sum> of irreducibles, proving the <Weyl complete reducibility theorem>.
A <derivation of a Lie algebra> is a linear map $D:\mathfrak g\to\mathfrak g$ satisfying the <Leibniz rule>
$$
D[x,y]=[Dx,y]+[x,Dy].
$$
The space $\operatorname{Der}(\mathfrak g)$ is a <vector subspace> of $\operatorname{End}(\mathfrak g)$. Equip it with the <commutator> $[D,E]=DE-ED$. Expanding the <Leibniz rule> twice gives
$$
\begin{aligned}
DE[x,y]&=[DEx,y]+[Ex,Dy]+[Dx,Ey]+[x,DEy],\\
ED[x,y]&=[EDx,y]+[Dx,Ey]+[Ex,Dy]+[x,EDy].
\end{aligned}
$$
Subtracting proves that $[D,E]$ is again a <derivation of a Lie algebra>. Antisymmetry and the <Jacobi identity> hold for the <commutator> in every associative endomorphism algebra, so this defines the <derivation Lie algebra>.
The <Jacobi identity> says that $\operatorname{ad}x$ is a <derivation of a Lie algebra>. Moreover, for $D\in\operatorname{Der}(\mathfrak g)$,
$$
[D,\operatorname{ad}x](y)=D[x,y]-[x,Dy]=[Dx,y],
$$
so
$$
\boxed{[D,\operatorname{ad}x]=\operatorname{ad}(Dx).}
$$
This makes $\operatorname{ad}\mathfrak g$ an <ideal of a Lie algebra> in $\operatorname{Der}(\mathfrak g)$.
Finally suppose $\mathfrak g$ is <semisimple Lie algebra>[semisimple]. Let $\mathfrak g$ act on $\operatorname{Der}(\mathfrak g)$ by $x\cdot D=[\operatorname{ad}x,D]$. The <Weyl complete reducibility theorem> supplies an invariant complement $M$ to $\operatorname{ad}\mathfrak g$. For $D\in M$, invariance gives $[\operatorname{ad}x,D]\in M$, while the <ideal> identity above gives $[\operatorname{ad}x,D]\in\operatorname{ad}\mathfrak g$. Their intersection is zero, so $\operatorname{ad}(Dx)=0$ for every $x$. The <center of a Lie algebra> of a <semisimple Lie algebra> is zero, hence $Dx=0$ for every $x$, and $M=0$. Therefore
$$
\boxed{\operatorname{Der}(\mathfrak g)=\operatorname{ad}\mathfrak g.}
$$
Every <derivation of a Lie algebra> is inner, and the element giving it is unique because the <center of a Lie algebra> vanishes.
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