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...

As the comments already mentioned, the claim is incorrect (The sequence itself is a subsequence for example). The flaw in your reasoning is in your recursive loop. You implicitly assume this loop will end in finite steps. This is by no means clear, since we can have infinitely many different subsequences
