Alternative Proof: If $(x_n)$ is bounded and diverges, then there exist two subsequences of $(x_n)$ that converge to different limits

Question

I know other proofs of this exist, but how can continue with this proof with $\epsilon$-neighborhoods?

Let $(x_n)$ be bounded but divergent. By the Bolzano–Weierstrass Theorem (BWT), $(x_n)$ contains a convergent subsequence, $(b_n)$.

Let $(b_n)\to L$. So, $\forall \epsilon \gt0 \exists N \in \mathbb{N}: \forall n \gt N |b_n -L| \lt \epsilon$. If there exists an $N'$ such that for all $n\gt N' |x_n -L| \lt \epsilon$, then $(x_n)$ converges - a contradiction.

Here is where I am stuck. Can I say there exist infinitely many points in the sequences outside the $\epsilon$-neighborhood without using the infinite extraction method found in other proofs for this?

Answer 1

$(x_n)$ converges to $L$ precisely when for all $\epsilon>0$ all but finitely many terms fall inside $B_\epsilon(L)$. For $(x_n)$ to not converge to $L$ it means there is some $\epsilon>0$ such that infinitely many terms of the sequence lie outside $B_\epsilon(L)$. Let us denote this subsequence by $(a_n)$. Now $(a_n)$ is also bounded and has a convergent subsequence $(y_n)$. Since $(y_n)$ is a sequence in the closed set $B_\epsilon(L)^c$, its limit must also be in the set. So it cannot be $L$.