Approaching elements in the image of a bounded operator with unbounded sequences In this question we showed that there exists a bounded operator $A:\ell_2\rightarrow \ell_2$ and a $y\in \overline{A(\ell_2)}$ with the following property: Every sequence $(x_n)_{n\in\mathbb{N}}$ in $\ell_2$ such that $A(x_n)\rightarrow y$ is unbounded. Does this hold more generally?


Question: Let $X, Y$ be Banach spaces and $A: X\rightarrow Y$ be a bounded operator such that $A(X)\neq \overline {A(X)}$. Does there always exist a $y\in \overline{A(X)}$ with the property that every sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ such that $A(x_n)\rightarrow y$, is unbounded? 
Equivalently, if $A(X)\subsetneq \overline {A(X)}$, is it always true that $\bigcup_{n\in\mathbb{N}}\overline{A(nB_X)} \subsetneq \overline{A(X)}$? 


In the last sentence, $B_X$ denotes the unit ball of $X$. 
 A: Yes, it's true that if $\bigcup_{n\in\mathbb{N}}\overline{A(nB_X)} = \overline{A(X)}$ then $A(X)=\overline{A(X)}$.  The proof is essentially the same as the proof of the Open Mapping Theorem, although I don't see a direct way to get the result from the OMT. 
Proof. First replace $Y$ by $\overline{A(X)}$ to simplify things (avoid talking about relative topology of $\overline{A(X)}$.) The assumption $\bigcup_{n\in\mathbb{N}}\overline{A(nB_X)} = Y$ and Baire's category theorem imply that there exists $n$ such that $\overline{A(nB_X)}$ has nonempty interior. Then the set $A(nB_X)-A(nB_X)$ contains $0$ in its interior. Since this set is contained in $A(2nB_X)$, we get there is $r>0$ such that 
$$
rB_Y\subset \overline{A(2nB_X)}
\tag1
$$
I claim that 
$$
\overline{A(2nB_X)} \subset A(4nB_X)
\tag2
$$
If (2) is proved, then $A(4nB_X)$ contains $rB_Y$, which implies 
$$ A(X) = \bigcup_{n=1}^\infty A(nB_X) = Y$$

Proof of (2): take any $y_1\in \overline{A(2nB_X)}$. There exists $x_1 \in 2nB_X$ such that $\|y_1-Ax_1\|<r/2$. Let $y_2=y_1-Ax_1$. By $(1)$, we have $y_2\in (r/2)B_Y\subset \overline{nB_X}$. So there exists $x_2\in nB_X$ such that $\|y_2-Ax_2\|<r/2^2$. This goes on, with
$$
y_{n+1}=y_n-Ax_n, \qquad x_n\in \frac{2n}{2^{n-1}}B_X
$$
Conclusion: $y_1=Ax$ where $x=\sum_{n=1}^\infty x_n\in B(4nB_X)$.
