Centralizer of a regular semisimple element is Cartan

${\mathfrak{g}}^x$ is a Lie subalgebra: for $y,z\in {\mathfrak{g}}^x$, Jacobi and [L5] give $[x,[y,z]]=[[x,y],z]+[y,[x,z]]=0$.

${\mathfrak{g}}^x$ equals its normalizer. Let $y\in N_{\mathfrak{g}}({\mathfrak{g}}^x)$. Since $x\in {\mathfrak{g}}^x$, we have $[y,x]\in {\mathfrak{g}}^x$, so $[x,[y,x]]=0$. Decompose $y={\sum }_{\lambda }{y}_{\lambda }$ in the eigenspaces of the semisimple operator ${\mathrm{ad}}_x$ [L2]. Then $[x,y]={\sum }_{\lambda }\lambda y_{\lambda }$ and $[x,[y,x]]=-{\sum }_{\lambda }{\lambda }^2{y}_{\lambda }=0$; the summands lie in distinct eigenspaces, so ${\lambda }^2{y}_{\lambda }=0$, and since the field is $\mathbb{C}$ we get $y_{\lambda }=0$ for $\lambda \mathrm{\ne }0$. Hence $y=y_0\in {\mathfrak{g}}^x$, so $N_{\mathfrak{g}}({\mathfrak{g}}^x)={\mathfrak{g}}^x$.

Every $z\in {\mathfrak{g}}^x$ acts by zero on ${\mathfrak{g}}^x$. Since $[x,z]=0$, the operators $A={\mathrm{ad}}_x$ and $B={\mathrm{ad}}_z$ commute [L2], and $A$ is semisimple [L2], so $\mathfrak{g}={⨁}_{\lambda }{V}_{\lambda }$ with $V_{\lambda }=\ker (A-\lambda )$ and each $V_{\lambda }$ is $B$-invariant. For $t\in \mathbb{C}$ put $M_t=A+tB={\mathrm{ad}}_{x+tz}$. For $t\mathrm{\ne }0$ an element $v={\sum }_{\lambda }{v}_{\lambda }$ with $v_{\lambda }\in V_{\lambda }$ is killed by $M_t$ exactly when $\lambda v_{\lambda }+tB{v}_{\lambda }=0$ for every $\lambda$, so $\dim \ker M_t=\dim \ker (B{\mathrm{\mid }}_{V_0})+{\sum }_{\lambda \mathrm{\ne }0}{m}_{\lambda }(t)$, where $m_{\lambda }(t)=\dim \ker (B+\frac{\lambda }{t}){\mathrm{\mid }}_{V_{\lambda }}$ is the multiplicity of the eigenvalue $-\frac{\lambda }{t}$ of the endomorphism $B{\mathrm{\mid }}_{V_{\lambda }}$ and therefore vanishes for all but finitely many $t$. Choosing $t$ outside this finite exceptional set and using [L1] at the element $x+tz$ gives $\dim \ker (B{\mathrm{\mid }}_{V_0})=\dim \ker M_t\ge m=\dim V_0$, hence $\ker (B{\mathrm{\mid }}_{V_0})=V_0$ and $B{\mathrm{\mid }}_{V_0}=0$; by definition $V_0={\mathfrak{g}}^x$.

Since $z\in {\mathfrak{g}}^x$ was arbitrary, step 1.3 shows that ${\mathrm{ad}}_z{\mathrm{\mid }}_{{\mathfrak{g}}^x}=0$ for every $z\in {\mathfrak{g}}^x$, so ${\mathfrak{g}}^x$ is abelian and in particular nilpotent; alternatively, every adjoint operator of ${\mathfrak{g}}^x$ is nilpotent on ${\mathfrak{g}}^x$ and [L4] applies. By step 1.2, ${\mathfrak{g}}^x$ equals its normalizer, so [L3] makes ${\mathfrak{g}}^x$ a Cartan subalgebra. When $\mathfrak{g}=0$ we have $x=0$, ${\mathfrak{g}}^x=0$, and the zero subalgebra is a Cartan subalgebra of the zero algebra; no nonempty choice occurs.