Cartan subalgebras of a direct sum

Write $R_i=\mathrm{rad}({\mathfrak{g}}_i)$ and $R=\mathrm{rad}(\mathfrak{g})$. The subspace $R_1\oplus R_2$ is a solvable ideal because brackets and every term of its derived series are computed componentwise, so $R_1\oplus R_2\subseteq R$. Conversely each projection ${\pi }_i(R)$ is an ideal of ${\mathfrak{g}}_i$ and is solvable: ${\pi }_i(R{)}^{(m)}={\pi }_i(R^{(m)})=0$ once $R^{(m)}=0$. Thus ${\pi }_i(R)\subseteq R_i$ and $R\subseteq R_1\oplus R_2$. Hence $R=R_1\oplus R_2=0$, proving that $\mathfrak{g}$ is semisimple before the Cartan/maximal-toral theorem is applied.

If each ${\mathfrak{h}}_i$ is a Cartan subalgebra of ${\mathfrak{g}}_i$, then ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ is nilpotent, being a direct sum of nilpotent algebras, and its normalizer is $N_{{\mathfrak{g}}_1}({\mathfrak{h}}_1)\oplus N_{{\mathfrak{g}}_2}({\mathfrak{h}}_2)={\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$: an element $x=x_1+x_2$ normalizes ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ exactly when $[x_i,{\mathfrak{h}}_i]\subseteq {\mathfrak{h}}_i$ for $i=1,2$, because brackets in a direct sum are computed componentwise and mixed brackets vanish.

Conversely let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. By step 1.1 and Cartan subalgebras are exactly maximal toral subalgebras it is maximal toral, hence abelian with all adjoint operators semisimple (Toral and maximal toral subalgebras). Let ${\mathfrak{h}}_i$ be the image of $\mathfrak{h}$ under the projection $\mathfrak{g}\to {\mathfrak{g}}_i$; each ${\mathfrak{h}}_i$ is abelian, since it is the image of an abelian subalgebra under a Lie-algebra homomorphism, and each of its elements is semisimple, because the adjoint operator of $x_1+x_2$ splits as the direct sum of the adjoint operators of $x_1$ and $x_2$, and a direct sum of endomorphisms is semisimple exactly when both summands are. Hence ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ is toral and contains $\mathfrak{h}$, so maximality gives $\mathfrak{h}={\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$.

Each ${\mathfrak{h}}_i$ is maximal toral in ${\mathfrak{g}}_i$: if ${\mathfrak{t}}_i⊋{\mathfrak{h}}_i$ were toral in ${\mathfrak{g}}_i$, then replacing the $i$th summand of ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ by ${\mathfrak{t}}_i$ would give a toral subalgebra of $\mathfrak{g}$ strictly containing $\mathfrak{h}$, contradicting maximality. By Cartan subalgebras are exactly maximal toral subalgebras each ${\mathfrak{h}}_i$ is a Cartan subalgebra of ${\mathfrak{g}}_i$, which completes the first half. The dimension formula is additivity of dimensions over a direct sum.

For the root statement, the eigenvectors of ${\mathrm{ad}}_H$ for $H=H_1+H_2\in \mathfrak{h}$ are exactly the sums of eigenvectors in the two summands: a functional on $\mathfrak{h}$ that is nonzero on both summands occurs for no nonzero eigenvector, while the roots of $\mathfrak{g}$ are the union of the roots of ${\mathfrak{g}}_1$ with respect to ${\mathfrak{h}}_1$ and of ${\mathfrak{g}}_2$ with respect to ${\mathfrak{h}}_2$, extended by zero on the other summand. Hence the root systems form a disjoint union, as asserted.