Proving that if $(a_n)$ is a bounded sequence and that if every convergent subsequence of $(a_n)$ converges to $a $, then $(a_n)$ converges to $a$. The proof that I found goes roughly like this : using the negation of the statement of convergence, it is possible to build a subsequence $(a_{n_k})$ that never enters the $\epsilon$-neighbourhood of $V_\epsilon(a)$. Then, since $(a_n)$ is bounded, this implies that $(a_{n_k})$ is also bounded and we can apply the Bolzano-Weierstrass theorem to it to obtain a subsequence of $(a_{n_k})$ which is also a subsequence of $(a_n)$. Since it is a subsequence of $(a_n)$, it is supposed to converge but since it is a subsequence of $(a_{n_k})$, it never enters the $\epsilon$-neighbourhood of $a$ and thus does not converge to $a$. This completes the proof.
My question is why should we use the Bolzano-Weierstrass theorem to produce a subsequence that never enters the $\epsilon$-neighbourhood of $a$ when we already constructed one namely $(a_{n_k})$. Should $(a_{n_k})$ not already be a subsequence of $(a_n)$ that does not converge to $a$ (and hence the proof is completed ?) ? I know there is a logical flaw in the argument (since the fact that $(a_n)$ was bounded was not used) but I cannot find it.
 A: No you got the rationale of the proof wrong.
Firstly, one assumes that $(a_n)$ does not converge. This yields the existence of some $\epsilon$ such that $\forall N\in \mathbb N$, there is some $n\geq N$ such that $a_n \notin V_{\epsilon}(a)$. This allows you to build a subsequence $a_{n_k}$ such that $\forall k, a_{n_k}\notin V_{\epsilon}(a)$.
Now, since $a_{n_k}$ is bounded, it has a convergence subsequence, say $b_n$, and $b_n$ is therefore a convergent subsequence of $a_{n_k}$, but also a convergent subsequence of $a_n$. Hence $b_n$ converges to $a$. But that is a contradiction, since $b_n$ is a subsequence of $a_{n_k}$.
A: In answer to your question, you initially have a subsequence which never enters the $\epsilon$ neighbourhood of $a$, but you do not know that the subsequence converges converges. So the proof uses boundedness and Bolzano-Weierstrass to construct a convergent subsequence which never enters the $\epsilon$ neighbourhood of $a$. Once you have such a convergent subsequence you can show that the limit is not $a$.
