[proofplan] We prove the two inclusions separately. The inclusion $H \subset C_L(H)$ follows from the abelianness of Cartan subalgebras in finite-dimensional semisimple Lie algebras over algebraically closed fields of characteristic zero. For the reverse inclusion, an element that centralizes $H$ automatically normalizes $H$, and the self-normalizing property of Cartan subalgebras forces it to lie in $H$. [/proofplan] [step:Use abelianness of $H$ to place $H$ inside its centralizer] Let $h_0 \in H$. Since $H$ is a Cartan subalgebra of the finite-dimensional semisimple Lie algebra $L$ over the algebraically closed field $k$ of characteristic zero, $H$ is abelian. Hence, for every $h \in H$, \begin{align*} [h_0,h]=0. \end{align*} By the definition of $C_L(H)$, this means $h_0 \in C_L(H)$. Since $h_0 \in H$ was arbitrary, we obtain \begin{align*} H \subset C_L(H). \end{align*} [/step] [step:Convert centralizing $H$ into normalizing $H$] Let $x \in C_L(H)$. By the definition of $C_L(H)$, for every $h \in H$, \begin{align*} [x,h]=0. \end{align*} Therefore \begin{align*} [x,h] \in H \end{align*} for every $h \in H$, because $0 \in H$. Thus \begin{align*} [x,H] \subset H. \end{align*} By the definition of the normalizer \begin{align*} N_L(H) := \{y \in L : [y,H] \subset H\}, \end{align*} we have $x \in N_L(H)$. [/step] [step:Apply the self-normalizing property of Cartan subalgebras] Since $H$ is a Cartan subalgebra of $L$, it is self-normalizing: \begin{align*} N_L(H)=H. \end{align*} The preceding step showed that every $x \in C_L(H)$ belongs to $N_L(H)$, and hence every $x \in C_L(H)$ belongs to $H$. Therefore \begin{align*} C_L(H) \subset H. \end{align*} [/step] [step:Combine the two inclusions] We have proved \begin{align*} H \subset C_L(H) \end{align*} and \begin{align*} C_L(H) \subset H. \end{align*} By antisymmetry of set inclusion, \begin{align*} C_L(H)=H. \end{align*} This is the desired equality. [/step]