Semisimple Lie algebras decompose into simple ideals

The assertion for $\mathfrak{g}=0$ is the empty direct sum. For nonzero $\mathfrak{g}$, if $\mathfrak{a}$ is any ideal, then $\mathfrak{a}\cap {\mathfrak{a}}^{\perp }$ is abelian. Indeed, for $u,v$ in that intersection and $z\in \mathfrak{g}$, invariance gives $K([u,v],z)=K(u,[v,z])=0$, because $[v,z]\in \mathfrak{a}$. By [L1], $[u,v]=0$. Semisimplicity makes this abelian ideal zero.

If $\mathfrak{g}\mathrm{\ne }0$, choose a nonzero ideal $\mathfrak{a}$ of least positive dimension; finite dimension makes this a choice from a finite set of integers. By step 1.1 and dimension, $\mathfrak{g}=\mathfrak{a}\oplus {\mathfrak{a}}^{\perp }$. Both summands are ideals, and their bracket lies in their intersection, hence is zero.

The minimal ideal $\mathfrak{a}$ is not abelian, because a nonzero abelian ideal is solvable. If $0\mathrm{\ne }\mathfrak{j}⊲\mathfrak{a}$, then $[{\mathfrak{a}}^{\perp },\mathfrak{j}]=0$ and $[\mathfrak{a},\mathfrak{j}]\subseteq \mathfrak{j}$, so $\mathfrak{j}$ is an ideal of $\mathfrak{g}$; minimality gives $\mathfrak{j}=\mathfrak{a}$. Thus $\mathfrak{a}$ is simple. Assume inductively that every semisimple algebra of smaller dimension has the asserted decomposition.

The complement ${\mathfrak{a}}^{\perp }$ is semisimple: any solvable ideal in it is, because the two summands commute, also a solvable ideal of $\mathfrak{g}$, and hence zero. Its dimension is smaller, so the induction hypothesis in step 3.1 decomposes it into finitely many simple ideals. Adjoining $\mathfrak{a}$ proves the result; a simple $\mathfrak{g}$ is the one-summand case.