= Solution
We prove closure using <complete reducibility of semisimple Lie algebra representations>, rather than assuming that a <Lie subalgebra> is closed under polynomials in its elements. Write the matrix <Jordan–Chevalley decomposition> as $X=S+N$. On $\operatorname{End}(V)$, the <commutator> operator $\operatorname{ad}S$ is diagonalizable: if $S$ has eigenspaces $V_\lambda$, then its eigenvalue on $\operatorname{Hom}(V_\lambda,V_\mu)$ is $\mu-\lambda$. The operator $\operatorname{ad}N$ is nilpotent: if $N^r=0$, then
$$
(\operatorname{ad}N)^k(A)=\sum_{j=0}^k(-1)^j\binom{k}{j}N^{k-j}AN^j=0\qquad(k\geq2r-1).
$$
The two operators commute because $[S,N]=0$. Uniqueness of the <Jordan–Chevalley decomposition> therefore says that $\operatorname{ad}S$ is the semisimple part of $\operatorname{ad}X$. Since a matrix's semisimple part is a polynomial in that matrix, and $\mathfrak g$ is invariant under $\operatorname{ad}X$, it follows that
$$
[S,\mathfrak g]\subseteq\mathfrak g.
$$
Now regard $\operatorname{End}(V)$ as a <Lie algebra representation> of $\mathfrak g$ under commutators. By <complete reducibility of semisimple Lie algebra representations>, there is an invariant complement $M$ with $\operatorname{End}(V)=\mathfrak g\oplus M$. Write $S=Y+C$, with $Y\in\mathfrak g$ and $C\in M$. For $A\in\mathfrak g$, the normalization property gives $[C,A]=[S,A]-[Y,A]\in\mathfrak g$, while invariance of $M$ gives $[C,A]\in M$. Thus $[C,A]=0$: $C$ commutes with the whole <Lie algebra>.
Decompose $V=\bigoplus_jV_j$ into irreducible $\mathfrak g$-modules. Each $V_j$ is preserved by $X$, hence by its polynomial parts $S,N$, and also by $Y$; therefore it is preserved by $C$. By the <Schur lemma>, $C|_{V_j}=c_jI$. A <semisimple Lie algebra> is a <perfect Lie algebra>, so every element is a sum of <Lie brackets> and has trace zero on every finite-dimensional representation. Consequently $\operatorname{tr}(X|_{V_j})=\operatorname{tr}(Y|_{V_j})=0$. Also $N|_{V_j}$ is nilpotent, so it has trace zero; hence
$$
\operatorname{tr}(S|_{V_j})=0,\qquad c_j\dim V_j=\operatorname{tr}(C|_{V_j})=0.
$$
Thus every $c_j$ is zero, $C=0$, and $S=Y\in\mathfrak g$. Finally $N=X-S\in\mathfrak g$. We have proved
$$
\boxed{X_s,X_n\in\mathfrak g.}
$$
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