$\overline{\{Ax:\|x\|\leq 1\}}\subset \{Ax:\|x\|\leq 2\}$ Continuing with the previous question, I want to ask one more question which seems true but I don't have any idea again. If $A\colon X\to \mathbb{C}^{n}$ is a finite rank bounded linear operator, then does the following holds?
$$\overline{\{Ax:\|x\|\leq 1\}}\subset \{Ax:\|x\|\leq 2\}$$
Without any further assumptions on $X$, I think this might be false, but I can't find any counter examples. It seems that $X=c_{0}$ given in the answer to the previous question can't be a counterexample. 
 A: The statement is true, and it is even true if you would replace the $2$ with $1+\varepsilon$.
First, we can consider $A$ as an operator from $X$ onto its range, which is again finite-dimensional.
So, without loss of generality we can assume that $A$ is surjective.
We define the set
$$
 D=\{Ax : \|x\|\leq 1\}.
$$
Using linearity, we can see that the original claim is equivalent to
$$
 \overline D \subset 2D.
$$
Note that it can be shown that $D$ is convex, symmetric, and $0$ is in the interior of $D$.
Let $\delta>0$ be such that $\overline{B_\delta(0)}\subset D$.
Consider a sequence $z_n\in D$ such that $z_n\to z$, but $z\not\in D$.
After some calculations, it can be seen that
$$
\gamma_n^{-1} z
=
(1-\beta_n) z_n
+
\beta_n\delta \|z-z_n\|^{-1} (z-z_n)
$$
with a suitable $\beta_n\in (0,1)$ and
where $\gamma_n= 1+\delta^{-1} \|z-z_n\|$.
Since this is a convex combination, this implies $\gamma_n^{-1}z \in D$,
or $z\in \gamma_n D$.
Since $\gamma_n\to 1$, this means that
$$
\overline D \subset (1+\varepsilon) D
$$
for every $\varepsilon>0$.
This is only a sketch of a proof and maybe there are less complicated ways to prove your claim.
