Let $a_1,a_2\dots$ be a bounded sequence in $\mathbb{R}$ that does not converge. Prove that the sequence has two subsequences that converge to different limits
Here is my proof: Let $s_n$ be a bounded sequence in $\mathbb{R}$ that does not converge. Then by the Bolzano Weierstrass theorem there exists a subsequence, $s_{n_k}$, that converges. Let $a$ be the limit of $s_{n_k}$. By definition $s_n$ does not converge so it can not converge to $a$. That is, there exists $\epsilon >0$ such that for all $N\in\mathbb{R}$, there exists $n>N$ such that $|s_n-a|\ge\epsilon$. So there exists a sequence $s_m$ that is a subsequence of $s_n$ with $m>N$. By its definition $s_m$ does not have a subsequence that converges to $a$. However, $s_m$ is bounded because $s_n$ is bounded. So by the Bolzano Weierstrass theorem there exists a subsequence, $s_{m_j}$, that converges. Clearly $s_{m_j}$ does not converge to $a$. $\square$
Am I correct in saying "By its definition $s_m$ does not have a subsequence that converges to $a$." This is what my proof relies on, but I'm not sure I have justified it correctly. Thanks in advance for any input!