Lemma: If $\mathfrak{k}$ is a subalgebra of $\mathfrak{g}$ that contains an Engel subalgebra, then $\mathfrak{k}$ is self-normalizing.
Proof: Suppose $\mathfrak{k}\supset \mathfrak{g}_0(ad\; x)$ for some $x\in \mathfrak{g}$, then $x\in\mathfrak{k}\subset N_{\mathfrak{g}}(\mathfrak{k})$ and $[x\mathfrak{k}]\subset \mathfrak{k}$. $ad\; x$ acts on $N_{\mathfrak{g}}(\mathfrak{k})/\mathfrak{k}$ and without eigenvalue 0 (?!!!). However $[x,N_{\mathfrak{g}}(\mathfrak{k})]\subset \mathfrak{k}$, so action is trivial. This means $N_{\mathfrak{g}}(\mathfrak{k}) = \mathfrak{k}$.
I understand everything from the proof except the part that ad $x$ has no 0 eigenvalue on $N_{\mathfrak{g}}(\mathfrak{k})/\mathfrak{k}$. Can someone please explain why?
$\mathfrak{g}_0(ad\ x)$ contains all $y$ such that $ad\ x (y)=0$. So the zero eigenspace of $ad\ x$ in $N_g(\mathfrak{k})$ is contained in $\mathfrak{k}$. Thus there is no zero eigenspace in the factor $N_g(\mathfrak{k})/\mathfrak{k}$