I know this post is the same question, but my proof is not detailed in that thread so I am here to check if it's valid.
Assume for a contradiction that there does not exist a value $M$ such that there are infinitely many $a_n > M$ and infinitely many $a_n < M$, this implies that $\forall\epsilon>0 \ \exists N \forall n>N \ |a_n-\text{inf}(\{a_n:n\in\mathbb{N}\}) < \epsilon$, this would contradict the assumption that $a_n$ does not converge, meaning $M$ does exist.
Define $m_n^+$ to be the subsequence constructed by $a_n>M$, and $m_n^-$ to be the subsequences constructed by $a_n<M$
Now, there are two possible cases, we will prove that both necessitate two subsequences:
(i) The convergent subsequences provided by Bolzano-Weierstrass for $a_n^+$ and $a_n^-$ are different, in which case, our proof is complete by the fact that the provided subsequences are also subsequences of $a_n$.
(ii) The convergent subsequences provided by Bolzano-Weierstrass are the same, in which case they both converge to $M$. Notice that if this is the case $a_n^+$ and $a_n^-$ cannot both be convergent, as if they were they would both converge to $M$, meaning $a_n$ would converge to $M$ which would contradict our assumption that $a_n$ does not converge. This means that atleast $a_n^+$ or $a_n^-$ does not converge, if it is $a_n^-$ that does not converge, apply the same logic from the start of the proof to find a value $K$ which there are infinitely many $a_n^-<K$ and infinitely many $a_n^->K$, now take the convergent subsequence provided by Bollzano-Weierstrass for the sequence $k_n$ defined by all the time where $a_n^-<K$ and notice that this cannot possibly converge to $M$ as M is not within the bounds of the supremem or infinum of the sequence $k_n$, our proof is complete by the fact that $k_n$ and $a_n^-$ converge to different values.
NOTE: in part (ii) if it is the case that $a_n^+$ does not converge, the same proof applies with identical logic, just a slight change in greater-than/lesser-than relations.