Why subsequence $ a_{n_{p}( k (\epsilon) )} $ doesn't converge to $ b $ if its indexes depend on $ \epsilon $? Lemma used for proof of a theorem:
$ L \in \mathbb{R} $ is a partial limit of sequence $ ( a_n ) $ iff  for all $ \epsilon > 0 $ the set $ \{ n : | a_n - L | < \epsilon \} $ is infinite.
Notation:  $ \mathscr{P}(a_n) $ is the set of all real partial limits of $ ( a_n) $
Theorem: Let $ ( a_n )  $ be a bounded sequence and let $ ( b_m) \subseteq \mathscr{P}(a_n)$ be a sequence of partial limits of $ ( a_n) $ s.t. $ \lim\limits_{m \to \infty} b_m = b$ .
Show that $ b \in \mathscr{P}(a_n)  $.
Proof( from class notes ) :  We have $ \lim\limits_{m \to \infty} b_m = b$, therefore for every $ \epsilon > 0 $ there exists $ b_{k(\epsilon)} $ s.t. $ | b_{k(\epsilon)} - b | < \epsilon/2 $ ( since  $ 
\forall \epsilon > 0$ there exists an index $k(\epsilon) $ s.t. $ \forall n>k(\epsilon) . $ $| b_n - b | < \epsilon $ , and I just chose one such index ), on the other hand $ b_{k(\epsilon)} \in \mathscr{P}(a_n)  $ ( meaning that $ b_{k(\epsilon)} $ is a partial limit ) so therefore there exists a subsequence $ a_{n_{p}( k (\epsilon) )} $ s.t. $ a_{n_{p}( k (\epsilon) )}  \xrightarrow{ p \to \infty }  b_{k(\epsilon)} $ , therefore there exists $ p_{0(k ( \epsilon) )} \in \mathbb{N} $ s.t.  for all $ p > p_{0(k ( \epsilon) )}  $ we have $  | a_{n_{p}( k (\epsilon) )}  - b_{k(\epsilon)} | < \epsilon/2  $.
So for all $ p > p_{0(k ( \epsilon) )}  $ we have $ | a_{n_{p}( k (\epsilon) )}  - b | \leq | a_{n_{p}( k (\epsilon) )}  - b_{k(\epsilon)} | + | b_{k(\epsilon)} - b | < \epsilon/2 + \epsilon/2 = \epsilon  $, meaning the set $ \{ n : | a_n - L | < \epsilon \} $ is infinite and therefore $ b \in \mathscr{P}(a_n)  $ . Q.E.D.
My question:
at the last steps of the proof, we got " $ p > p_{0(k ( \epsilon) )}. | a_{n_{p}( k (\epsilon) )}  - b | < \epsilon  $  " and my professor said that it is false to think that we got here a subsequence $ a_{n_{p}( k (\epsilon) )} $ that converges to $ b$ and that is because the indexes in a subsequence that converges to $ b $ should not depend on $ \epsilon $ ( And in $ a_{n_{p}( k (\epsilon) )} $ the indexes $ n_{p}( k (\epsilon) ) $ depend on $ \epsilon $  ).
So I didn't really understand this ( I asked him for further clarification but the answer was overall the same and I still didn't understand. ), why the subsequence  $ a_{n_{p}( k (\epsilon) )} $ does not converge to $ b $? so what if the indexes depend on $ \epsilon $ ?
 A: Consider the sequence $\{a_n\}_{n=1}^\infty$ where
\begin{aligned}
a_n=\begin{cases}1 & \text{if }n \neq p^k \text{ for some prime } p,\, k \in \mathbb{N}\\
p^{-1}-k^{-\frac{p+1}{p}} & \text{if }n = p^k \text{ for some prime } p,\, k \in \mathbb{N}\end{cases}
\end{aligned}
So $\lim_{k \to \infty} a_{p^k}=\lim_{k \to \infty} p^{-1}-k^{-\frac{p+1}{p}}=p^{-1}$ and $\lim_{m \to \infty} p_m^{-1}=0$ where $p_m$ denotes the $m$th prime number. Think about what issue might arise in applying your argument to the above example.
Edit:
I am annoyed by my first counterexample, so I want to also consider the sequence $\{b_n\}_{n=1}^\infty$ where
\begin{aligned}
b_n=\begin{cases}1 & \text{if }n \neq p^k \text{ for some prime } p,\, k \in \mathbb{N}\\
p^{-1}-pk^{-\frac{p+1}{p}} & \text{if }n = p^k \text{ for some prime } p,\, k \in \mathbb{N}\end{cases}
\end{aligned}
(I would like to think that there are in some sense better counterexamples to illustrate the problem with the proposed argument but I am a one mediocre idea kind of guy)
