Assume $(a_n)$ is a bounded sequence with the property that every convergent subsequence of converges to the same limit $a\in\mathbb{R}$

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 have tried to prove this by contradiction. That is, assume $$(a_n)$$ does not converge to $$a.$$ Then there are two cases:

1. $$(a_n)$$ converges to some other value $$b.$$
2. $$(a_n)$$ diverges.

For case1, if $$(a_n)$$ converges to $$b,$$ then all the subsequences of $$(a_n)$$ should converge to $$b,$$ which contradicts that all convergent subsequences converge to $$a.$$

For case2, if $$(a_n)$$ diverges, then I can find counterexamples to show $$(a_n)$$ does not have the property that all convergent subsequences converge to the same limit. For example $$(-1)^n$$ has two subsequences with different limits. $$(a_n)$$

Then I figured out that I did not use the hypothesis that $$(a_n)$$ is bounded. So there might be something wrong in my proof, but I could not find it out.

If $$(a_n)$$ does not converge to $$a$$ then there exists $$\epsilon >0$$ and integers $$n_k$$ such that $$n_1 and $$|a_{n_k}-a| \geq \epsilon$$ for all $$k$$. Now $$(a_{n_k})$$ is bounded sequence and hence it has a convergent subsequence. But the inequality $$|a_{n_k}-a| \geq \epsilon$$ holds for this subsequence also. Do you see a contradiction now?
• Case 2) in your answer does not make sense. $(a_n)$ is a given sequence and an example like $(-1)^{n}$ has no role in this result. @Howard Mar 21 '21 at 5:38