inequality for compact contained banach space

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.