Bounded linear operator maps norm-bounded, closed sets to closed sets. Implies closed range? Question
Suppose $T:X\rightarrow Y$ is a continuous, injective linear operator between Banach spaces. Suppose, in addition, that $T$ maps norm bounded closed sets in $X$ to closed sets in $Y$. Then the range of $T$ is closed in $Y$.
This is a problem related to one given An Invitation to Operator Theory by Abramovich and Aliprantis and I'd just like to verify my proof.
Attempt
We assume that $T$ is as above, and we shall prove that it has closed range. If $y_n = T x_n$ and $y_n\rightarrow y$, we want to show that $y=Tx$ for some $x\in X$. First, suppose $\{x_n\}_{n\geq 1}$ is unbounded. Then
$$\lim_n\, T(x_n/\|x_n\|) = \lim_n\, y_n/\|x_n\| = 0.$$
But the set $B=\{ x\in X: \| x\|=1\}$ is closed and norm-bounded, so its image under $T$ is closed. In particular, we must have $Tz=0$ for some $z\in B$. This contradicts the fact that $T$ is injective. So the sequence $\{x_n\}_{n\geq 1}$ is bounded in $X$. Since $\{x_n\}_{n\geq 1}$ is bounded, the set
$$ A=\mathrm{cl} \{ x_1, x_2, \ldots, x_n, \ldots \}$$
is closed and norm bounded. Hence $T(A)$ is closed in $Y$. In particular, $y=\lim_n\, Tx_n = Tx$ for some $x\in A \subset X$. So the range of $T$ is closed.
Thanks in advance!
 A: Your proof looks good.  
Here's another possible line you could follow.

Lemma. Suppose $X,Y$ are Banach spaces, $T : X \to Y$ is continuous and injective.  Then the range of $T$ is closed if and only if there is a constant $c > 0$ such that $\|Tx\| \ge c\|x\|$ for all $x\in X$ (we say such $T$ is bounded below).

Proof.  For the forward direction, use the open mapping theorem.  (But we don't actually need the forward direction for this problem).  For the reverse direction, if $T$ is bounded below then $T^{-1}$ is bounded.  So $TX = (T^{-1})^{-1} X$ is closed, being the preimage of a closed set under a continuous map.
Now for the problem: let $S$ be the unit sphere of $X$.  By assumption $TS$ is closed, and by the injectivity of $T$ it does not contain 0.  Hence its complement contains an open ball of some radius $c$ about 0.  This means that $\|Tx\| \ge c\|x\|$ for all $x \in S$, and by linearity the same holds for all $x \in X$.  So $T$ is bounded below, and by our lemma it has closed range.
A: You are assuming that $(x_{n})$ unbounded implies $\left(\frac{1}{\|x_{n}\|}\right)$ goes to $0$, so that you can then conclude $\lim_{n\to\infty}\frac{y_{n}}{\|x_{n}\|}=0$.
This is not necessarily true.
For example, the sequence $(x_{n})$ in $\mathbb{R}$ given by $x_{n}=\begin{cases}n & \mbox{if }n\mbox{ is even}\\1 & \mbox{if }n\mbox{ is odd}\end{cases}$ is unbounded.
However, $\frac{1}{\|x_{n}\|}=\begin{cases}\frac{1}{n} & \mbox{if }n\mbox{ is even}\\1 & \mbox{if }n\mbox{ is odd}\end{cases}$ does not converge.
Fortunately, $\left(\frac{1}{\|x_{n}\|}\right)$ does contain a subsequence converging to $0$. So you can get around the problem with a minor change to your proof:
Assume $(x_{n})$ is not bounded. Then there is a subsequence $(x_{n_{k}})$ of $(x_{n})$ such that $x_{n_{k}}\not=0$ for all $k$ and $\lim_{k\to\infty}\|x_{n_{k}}\|=\infty$.
Then $\lim_{k\to\infty}\frac{y_{n_{k}}}{\|x_{n_{k}}\|}=0$.
