Let $S$ be a vector subspace of the Lie algebra $\mathfrak{g}$.
Is $S$ an ideal of the normalizer $N_\mathfrak{g}(S)$, I would say yes since:
$\forall n\in N_\mathfrak{g}(S), [n,S]\subseteq S\implies [N_\mathfrak{g}(S),S]\subseteq S$ which is the definition of an ideal right?
The normalizer of a subspace $S$ in $\mathfrak{g}$ is the largest subalgebra of $\mathfrak{g}$ containing $S$ as an ideal. Here "containing" means that $S\subseteq N_{\mathfrak{g}}(S)$, i.e., that $[S,S]\subseteq S$. So $S$ is a subalgebra, and of course then also an ideal, as you have shown.