If $X \subset Y\subset Z$ then for $\epsilon>0$ there exists $c(\epsilon)$ such that $\|x\|_Y\leq \epsilon\|x\|_X+c\|x\|_Z$ Problem

Let $(X,\|\cdot\|_X), (Y,\|\cdot\|_Y), (Z,\|\cdot\|_Z)$ be Banach spaces. Suppose $X\subset Y\subset Z$, the inclusion $X$ into $Y$ is bounded and compact and the inclusion $Y\subset Z$ is bounded. That is, $X$ is compactly imbedded in $Y$ and $Y$ is continuously imbedded into $Z$. Show that for any $\epsilon>0$, there exists $c=c(\epsilon)>0$ such that $$\|x\|_Y\leq \epsilon\|x\|_X+c\|x\|_Z$$ for all $x\in X$.

Attempt
I am hoping for a hint to help start this problem.
 A: Here is a proof:
Result: Let $X,Y,Z$ be Banach, $X\stackrel{i}{\hookrightarrow} Y\stackrel{j}{\hookrightarrow} Z$ be injections such that $i\in \operatorname{Hom}(X,Y)$ is compact and $j\in\operatorname{Hom}(Y,Z)$. Then
$$\forall \epsilon>0,\exists C>0,\forall x\in X: \Vert i(x)\Vert_Y\leq \epsilon\Vert x\Vert_X+C\Vert j\circ i(x)\Vert_Z.$$
Proof: Suppose not. Then
$$\exists \epsilon_0>0,\forall n\geq1,\exists x_n\in X: \Vert i(x_n)\Vert_Y> \epsilon_0\Vert x_n\Vert_X+n\Vert j\circ i(x_n)\Vert_Z.$$
Observe that $x_n\neq 0$ FIM $n$, for otherwise large enough $n$ provides $0>0,{\bf\large\unicode{x21af}}$. Thus we may assume, by switching to a subsequence and normalizing if necessary, that $\forall n: \Vert i(x_n)\Vert_Y=1$. Then
$$\forall n: \dfrac{1}{\epsilon_0}>\Vert x_n\Vert_X$$
and
$$\forall n:  \dfrac{1}{n}>\Vert j\circ i(x_n)\Vert_Z,$$
so that $\{x_n\}_n\subseteq X$ is bounded and $j\circ i(x_n)\to0_Z$. Since $i$ is compact $\exists n_k\subseteq n,\exists y\in Y:i(x_{n_k})\to y$. Since $j$ is continuous $j\circ i(x_{n_k}) \to j(y)$. We also have $j\circ i(x_{n_k})\to 0_Z$, so that $j(y)=0_Z$, and consequently $y=0_Y$ by the injectivity of $j$.
Since $\left\vert \Vert i(x_{n_k})\Vert_Y-\Vert y\Vert_Y \right\vert \leq \Vert i(x_{n_k})-y\Vert_Y\to 0$,
$$1=\Vert i(x_{n_k})\Vert_Y\to \Vert y\Vert_Y=0 \implies 1=0,{\bf\large\unicode{x21af}}.\checkmark$$
