About a proof of Bolzano-Weierstrass theorem Recently I learned about the Bolzano-Weierstrass theorem. The theorem is the following:
In $\mathbb R$ every bounded sequence contains a convergent subsequence. 
A sequence $a_n$ is bounded if $a_n \in [-C,C]$ for some $C$. The proof that I saw was doing a bisection of $[-C,C]$ into subintervals of decreasing length. I understand the proof but I found another proof that seems slightly less complicated but I don't know if it is correct. Please can someone read my proof and tell me if it is correct? 
Proof: If $a_n$ is bounded then $A = \{a_n \mid n \in \mathbb N \}$ is bounded. By the axiom of completeness $a = \sup A$ exists. By the definition of $\sup$ for every $k$ there is $a_{n_k} \in A$ with $|a_{n_k} - a|<{1 \over k}$. The $a_{n_k}$ are a convergent subsequence of $a_n$.
 A: If you take the sequence $(-2,2,-1,1,-1/2,1/2,-1/4,1/4,...)$ your argument using the supremum completely breaks down. However, you can rescue your approach by considering the limes superior, for instance in the form of
$$s=\limsup_{n\to\infty} a_n=\lim_{n\to\infty}\sup_{k\ge n}a_k=\inf_{n\in\mathbb N}\sup_{k\ge n}a_k.$$
For $\varepsilon_j=2^{-j}>0$ and $N_j\in\mathbb N$ you will find an index $k_j\ge N_j$ with $s-ε<a_{k_j}\le s$. Then find the next index with $N_{j+1}=k_j+1$.
A: My mistake was that I assumed there are infinitely many $a_n$ near the $\sup$ of the set. But this is not implied by the definition of $\sup$. 
A: The completeness of the real line and Bolzano-Weierstrass theorem are actually equivalent hence invoking the former to prove the latter is questionable.
A: The problem with your argument is that you cannot state that $a_{n_k}$ is a convergent sequence.
(Hint: it may not even be a sequence in $(a_{n})$).
All you know is that for every $k$ there is an element $somewhere$ in the sequence $(a_{n})$ that is in a neighbourhood of size ${1 \over k}$ of $a$.
However your argument can be made correct if you identify the property such elements $a_{n_k}$ must have in order to create a convergent sequence, and show that such a sequence exists.
Hint: When is a bounded sequence necessarily convergent?
