inequality for compact contained banach space Let $X_0$,$X_1$ and $X_2$ be Banach spaces such that
$$X_2\subset\subset X_1\subset X_0$$
I want to prove that for every $\epsilon>0$, there exists a constant $C_\epsilon>0$ such that for all $x\in X_2$, the inequality
$$\lVert x\rVert_{X_1}\leq\epsilon\lVert x\rVert_{X_2}+C_\epsilon \lVert x\rVert_{X_0}$$
holds
My attempt is to argue by contradiction i.e. suppose that there is $\epsilon>0$ such that for all $C>0$, there is an $x_C\in X_2$ such that $$\lVert x\rVert_{X_1}>\epsilon\lVert x\rVert_{X_2}+C \lVert x\rVert_{X_0}$$
then let $C=k\in\mathbb{N}$, it gives a sequence $\{x_k\}\subset X_2$ satisfying this inequality with $C=k$.
divided both side $\frac{1}{\lVert x\rVert_{X_1}}$ then we can find a bounded sequence in $X_2$ that is $$y_k=\frac{x_k}{\lVert x\rVert_{X_1}}$$
and $y_k$ satisfies $$1>\epsilon\lVert y_k\rVert_{X_2}+k \lVert y_k\rVert_{X_0}$$
so it should have a subsequence which is convergent both in $X_1$ and $X_0$, denoted by $\{z_k\}$
since $X_1$ is Banach so it has a limit point say $z\in X_1$
then I don't know how to continue.
My teacher explains the notions he used
$$X_2\subset\subset X_1\subset X_0$$ here means the embedding of $X_2$ into $X_1$ is compact and the embedding of $X_0$ is continuous. In particular, every bounded sequence in $X_2$ has a subsequence which is convergent in $X_1$ and any convergent sequence in $X_1$ is also convergent in $X_0$.
 A: I think you are almost done so I will just follow your steps.
assume contrary which means $\exists\epsilon>0$ $\forall C>0$ $\exists x_C\in X_2$ such that $$\lVert x_C\rVert_{X_1}>\epsilon\lVert x_C\rVert_{X_2}+C\lVert x_C\rVert_{X_0}$$
let $C=k\in\mathbb{N}$ and $x_k$ be the correspnding $x_C$
then $\{x_k\}$ is a sequence in $X_2$ satisfying  $$\lVert x_k\rVert_{X_1}>\epsilon\lVert x_k\rVert_{X_2}+k\lVert x_k\rVert_{X_0}$$
then for each $x_k$ we have $$\lVert x_k\rVert_{X_1}>\epsilon\lVert x_k\rVert_{X_2}+k\lVert x_k\rVert_{X_0}\geq 0+0=0$$
so $\lVert x_k\rVert_{X_1}\neq0$
hence we can define $y_k=\frac{x_k}{\lVert x_k\rVert_{X_1}}$
then $$\frac{1}{\lVert x_k\rVert_{X_1}}\lVert x_k\rVert_{X_1}>\epsilon\frac{1}{\lVert x_k\rVert_{X_1}}\lVert x_k\rVert_{X_2}+k\frac{1}{\lVert x_k\rVert_{X_1}}\lVert x_k\rVert_{X_0}$$
which implies $$1>\epsilon\lVert y_k\rVert_{X_2}+k\lVert y_k\rVert_{X_0}$$
so $\{y_k\}$ is a bounded sequence in $X_2$
so it has a subsequence which is convergent in $X_1$
write $\{z_k\}$ as the convergent subsequence
and since $X_1$ is Banach, this sequence has a limit point say $z\in X_1$
then $\{z_k\}$ should also convergent to $z$ in $X_0$
since we have $$1>\epsilon\lVert z_k\rVert_{X_2}+k\lVert z_k\rVert_{X_0}>k\lVert z_k\rVert_{X_0}$$
so $$\lVert z_k\rVert_{X_0}<\frac{1}{k}$$
$\lVert z\rVert_{X_0}=\lim\limits_{k\to\infty}\lVert z_k\rVert_{X_0}=0$ implying $z=0$
but by the construction of $\{y_k\}$ we know that $\lVert z_k\rVert_{X_1}\equiv1$
which leads to contradiction.
