Given a bounded sequence $a_n$, show that if every convergent subsequence $a_{n_k} \rightarrow L$, then $a_n \rightarrow L$

Question

I am trying to solve Abbot real analysis 2nd edition 2.5.5. As per the title:

• Given a bounded sequence $a_n$, show that if every convergent subsequence converges to the same limit $L$, then $a_n \rightarrow L$.

I am studying it on my own and I'm not sure if what I did is correct. Here is what I tried:

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Let $\epsilon > 0$ be an arbirary number. Suppose $a_n$ does not converge to $L$, then $\forall N, \exists n>N \Rightarrow a_n>\epsilon$.

Choose $N$ such that for every convergent subsequence $a_{n_k}$ $n_k>N \Rightarrow |a_{n_k} - L| < \epsilon$ (can I do this??).

Let $S = \{a_k: k > N , |a_k - L| >= \epsilon\}$.

Lemma: $S$ is empty (if I can proove this, what follows is trivial)

proof (by contradiction). Let $b_n$ be the non-empty sequence of elements of $S$. Since $a_n$ is bounded, then so is $b_n$. Therefore, by the Bolzano-Weierstrass theorem, $b_n$ has a convergent subsequence $b_{n_k}$, since $b_{n_k}$ is also a convergent subsequence of $a_n$, then $b_{n_k} \rightarrow L$. However, by construction of $S$, $\forall b_n, |b_n - L| >= \epsilon$, therefore $b_{n_k}$ cannot converge to $L$. Therefore $S$ is empty.

Any corrections more than welcome