Does every bounded, divergent sequence contain only convergent subsequences with at least two different limits? Claim: Every bounded and divergent sequence $(a_n)$ contains only convergent subsequences.
A bounded, divergent sequence $(a_n)$ implies that there is at least one convergent subsequence (by Bolzano Weierstrass) $(a_{nk}) \rightarrow l$. Now, because $(a_n)$ diverges, there exists some subsequence $(a_{jk})$ which isn't in the  $\epsilon -$ neighborhood of $l$. This means that either $(a_{jk})$ is divergent or converges to some other point (as it cannot be unbounded). If it converges to some other point $(a_{jk}) \rightarrow k$ we have shown that there exist two limit points and we are done. If, however, $(a_{jk})$ is divergent itself, I can apply Bolzano Weierstrass again and basically have a recursive loop until I am only left with only convergent subsequences that all converge to a different limit.
Is this valid? I'm not really sure if this is sufficient or even correct...

No. Consider the bounded sequence $1,2,3,1,2,3,1,2,3,\cdots $. One of its proper subsequences is $1,2,1,2,1,2,\cdots$ and it does not converge.
