I am trying to solve this exercise. Following the hint I am supposing that
$$\exists \varepsilon : \forall n\in\mathbb{N} \ \ \ \exists u_n\in X$$ with $$||u_n||_Y>\varepsilon||u_n||_X+n||u_n||_Z.$$
Now I would like to prove that the sequence $(u_n)$ is bounded in $X$ in order to extrat a subsequence converging in $Y$ (and so in $Z$) and then...I sincerely don't know.
It could be a right way?
You can assume the $u_n$ are bounded in $X$, because with $x_n:=\frac{u_n}{\|u_n\|_X}$ (all $u_n$ must differ from zero) you get the same inequality by homogeneity of the norms: $$\|x_n\|_Y>\epsilon\|x_n\|_X +n\|x_n\|_Z.$$ Now $x_n$ must admit a converging subsequence to some $x\in Y$ by the compactness of the embedding of $X$ in $Y$. So just assume the sequence converges. From continuity of the embedding $Y\to Z$ you find that $x_n\to x$ also in $Z$. But also: $$n\|x_n\|_Z < \|x_n\|_Y-\epsilon\|x_n\|_X<\|x_n\|_Y\implies \|x_n\|_Z <\frac{\|x_n\|_Y}n.$$ the limit on the right is $0$ and thus $\|x_n\|_Z\to0$ and $x_n\to0$ in $Z$. It follows $x=0$.
So $\|x_n\|_Y\to0$. But it must always be bigger than $\epsilon\|x_n\|_X=\epsilon$. This is impossible.