Let $G$ be a (connected) semisimple algebraic group over an algebraically closed field $k$ of characteristic 0. We consider the adjoint representation $$ {\rm Ad}\colon G\to {\rm GL}({\mathfrak g}),$$ where ${\mathfrak g}={\rm Lie}\ G$.
I am looking for a reference to a proof of the following assertion:
Proposition. For any semisimple element $X\in{\mathfrak g}$, its stabilizer ${\rm Stab}_G(X)\subset G$ with respect to the adjoint representation is connected.
I think that I can prove the proposition; see my proof below.
Proof. Our semisimple element $X$ is contained in a Cartan subalgebra ${\mathfrak t}$ of ${\mathfrak g}$, which is the Lie algebra of a maximal torus $T$ of $G$. Then ${\mathfrak t}$ is an algebraic subalgebra of ${\mathfrak g}$. Let ${\langle X\rangle_{\rm alg}}\subset {\mathfrak g}$ denote the smallest algebraic subalgebra of ${\mathfrak g}$ containing $X$; then ${\langle X\rangle_{\rm alg}}\subseteq {\mathfrak t}$. It follows that ${\langle X\rangle_{\rm alg}}={\rm Lie}\ S$ for some subtorus $S\subseteq T$. Now (in characteristic 0) we have $${\rm Stab}_G(X)=\bigcap_{Y\in {\langle X\rangle_{\rm alg}}} {\rm Stab}_G(Y)=C_G(S),$$ where $C_G(S)$ denotes the centralizer of the torus $S$ in $G$. By Theorem 22.3 of Humphreys' book "Linear Algebraic Groups", $C_G(S)$ is connected, as required.
Edit: A similar argument shows that for any commutative subalgebra ${\mathfrak a}\subset {\mathfrak g}$ consisting of semisimple elements, its centralizer in $G$ $$ C_G({\mathfrak a}):=\bigcap_{X\in {\mathfrak a}} {\rm Stab}_G(X)$$ is connected (because the "algebraic closure" $\langle {\mathfrak a}\rangle_{\rm alg}$ of $\mathfrak a$ is the Lie algebra of some torus $S\subset G$).
