Bounded sequence with subsequences all converging to the same limit means that the sequence itself converges to the same limit. I have a question regarding one exercise in Stephen Abbotts' Understanding Analysis. The question is: Assume $(a_n)$ is a bounded sequence with the property that every convergent subsequence of $(a_n)$ converges to the same limit $a \in \mathbb{R}$. Show that $(a_n)$ must converge to $a$.
I was then presented this proof (from the solutions manual): Assume for contradiction that $(a_n)$ does not converge to $a$, then from the negation of convergence we have (I'm not quite sure in this part):
$$\exists \epsilon>0, \forall N\in \mathbb{N}, \exists n\in \mathbb{N} : n\geq N \wedge |a_n-a|\geq \epsilon. $$
Using this we can construct a subsequence $(a_{n_j})$ that diverges from $a$ as follows: for arbitrary $N\in \mathbb{N}$, we can always find a, say, $n_1$ such that, $n_1 \geq N$ where $|a_{n_1}-a|\geq \epsilon$. Since $n_1$ itself is in $\mathbb{N}$, then we can again find a $n_2 \geq n_1$ so that $|a_{n_2}-a|\geq \epsilon$. And in general we can find a $a_{n_{j+1}}$ after choosing an appropriate $a_{n_j}$ so that $|a_{n_{j+1}}-a|\geq \epsilon$.
From the construction of such a subsequence, isn't the contradiction that $(a_{n_j})$ does not converge to $a$ already proof that $(a_n)$ should converge to $a$? I am asking because the proof that I've read continues to say that the Bolzanno-Weierstrass theorem can be used to get another subsequence from the constructed $(a_{n_j})$ which diverges from $a$, and that the proof ends there. I think I understand it, but I fail to see why it is necessary. Can somebody enlighten me on this?

If you want to do it from scratch, argue as follows:
If $\left \{ a_{n} \right \}$ does not converge to $a$, the there is an $\epsilon >0$ and a subsequence $\left \{ a_{n_{k}} \right \}$ such that for each $k>0$, $\vert a_{n_{k}} \vert >\epsilon$. 
But $a_{n_{k}}\subseteq  a_{n}$, so it is bounded because $a_{n}$ is. Therefore, by Bolzano-Weierstrass, it contains a convergent subsequence, say $a_{n_{k_{l}}}$ which cannot converge to $a$. 
And now you have your contradiction. 
