Prove: the inverse of an absolutely continuous operator $A:E_1 \to E_2$ need not be continuous. I'm studying about statistical learning theory and I bumped into the following statement in my study material: 

Claim:  
If a linear normed space $E_1$ contains bounded
  noncompact sets, then the inverse operator $A^{-1}$ for an absolutely
  continuous operator $A:E_1\rightarrow E_2$ ($E_2$ also linear normed
  space) need not be continuous. 
Proof: 
Consider a bounded noncompact set in linear normed space
  $E_1$. Select in this set an infinite sequence:
$$f_1, f_2, ...,\;(||f_j||_{E_1} \leq c \in
 \mathbb{R})\;\;\;\;\;\;\;\;\;\;\;\;(1)$$  such that no subsequence of
  it is convergent. An infinite sequence  $$Af_1, Af_2, ...,\;\;\;\;\;\;\;\;\;\;\;\;(2)$$  from which convergent subsequence
  may be selected (since $A$ is absolutely continuous) corresponds in
  $E_2$ to this sequence. If the operator $A^{-1}$ were continuous, then
  a convergent subsequence 
$$f_{k}, f_{k+1}, ...,\;\;\;\;\;\;\;\;\;\;\;\;(3)$$
could be extracted from $(1)$ in space $E_1$. This however contradicts
  the choice of sequence $(1)$.

The above was somewhat clear to me, but I wasn't 100% sure, so I wanted to try prove this claim myself and then verify here if I understood it or not. 
UPDATE: I have added my own attempt as a possible answer.
My question was: is my proof valid? Or did I misunderstand something? 
 A: 
Claim:
If a linear normed space $E_1$ contains bounded noncompact sets, then the
  inverse operator $A^{−1}$ for an absolutely continuous one-to-one operator $A:E_1→E_2$ ($E_2$
  also linear normed space) need not be continuous.

Proof: 
Denote $\rho_k := ||f_k-f_{k+1}||_{E_1}$ and $r_k :=  ||Af_k-Af_{k+1}||_{E_2}$. Since $A$ is absolutely continuous, for arbitrary $\delta >0$ we can always find $\varepsilon>0$ for which: 
$$\sum_{k} r_k <\delta \;\;\;\text{when}\;\;\; \sum_{k}\rho_k  <\varepsilon,$$
for some subsequence $f_k, f_{k+1}, ...$ of $(1)$. That is, $r_k \to 0$ and $A f_k \to A f_0$, so $(2)$ converges to $Af_0$. In other words, due to absolutely continuity of $A$ we can always select a subsequence from $(1)$ which has a convergent image. 
Lets now assume that we have a continuous inverse operator
 $A^{-1}$. This means that for arbitrary $k$ and $\delta_1>0 $, we can always find a
 $\varepsilon_1>0$ for which:
$$ \rho_k <\delta_1 \;\;\;\text{when}\;\;\;r_k  <\varepsilon_1, $$
which means that $(1)$ has a convergent subsequence. But this is a
 contradiction since $(1)$ is an infinite divergent sequence $\blacksquare$
