Can someone please check my proof of the following "evident" equivalence stated in Kurzweil and Stellmacher.
A finite group $G$ is nilpotent if and only if $$U<N(U) \text{ whenever } U<G.$$
My attempt: First suppose that $G$ is nilpotent. Then by definition, every subgroup is subnormal. Let $U<G$. By subnormality, we can find $U_1,\ldots, U_d\leq G$ such that $$U=U_1\triangleleft U_2\triangleleft\ldots \triangleleft U_d=G.$$ Now because $U\neq G$, there exists $y_d\in G=U_d\setminus U$. Therefore $y_d^{-1} U_{d-1} y_d=U_{d-1}$. Similarly, there exists $y_{d-1}\in U_{d-1}\setminus U_{d-1}$ such that $y_{d-1}^{-1} U_{d-2} y_{d-1}=U_{d-2}$. Continuing this way, there exists $y_2\in U_2\setminus U_1=U$ such that $y_2^{-1}Uy_2=U$ and so $y_2\in N(U)\setminus U$, which completes this part.
Conversely, assume that $$U<N(U) \text{ whenever } U<G.$$ Since every subgroup is normal in its normaliser, we have $U\triangleleft N(U)$. If $N(U)=G$, we're done. Otherwise, $$U\triangleleft N(U)\triangleleft N(N(U)).$$ Continuing this way, we get $$U\triangleleft N(U)\triangleleft N(N(U))\triangleleft \ldots \triangleleft G$$ and this series must be finite because $G$ is finite.