Double Subsequences Suppose that $\{a_{n}\}$ and $\{b_{n}\}$ are bounded. Prove that $\{a_{n}b_{n}\}$ has  a convergent subsequence. 
In class this is how my professor argued:
By the Bolzano-Weierstrass Theorem, there exists a subsequence $\{a_{n_k}\}$ that converges to $a$. Since $\{b_n\}$ is bounded, $\{b_{n_k}\}$ is also bounded. So by the Bolzano-Weierstrass Theorem, there exists a subsequence of $\{b_{n_k}\}$ namely $\{b_{n_{{k_j}}}\}$ such that $\{b_{n_{{k_j}}}\}$ converges to $b$. 
In particular $\{a_{n_{{k_j}}}\}$ will converge to $a$. And note that $\{a_{n_{{k_j}}}b_{n_{{k_j}}}\}$ is a subsequence of $\{a_{n}b_{n}\}$. So $a_{n_{{k_j}}}b_{n_{{k_j}}} \to ab$. 
My question is why do we have to use so many subsequences. Is it wrong to argue as follows?
$\{a_{n}\},\{ b_{n} \}$ are both bounded, so by the Bolzano-Weierstrass Theorem, both sequences have a convergent subsequence. Namely $a_{n_k} \to a$ and $b_{n_k} \to b$. Then note that $\{a_{n_k}b_{n_k}\}$ is a subsequence of $\{a_{n}b_{n}\}$ which converges to $ab$. And we are done.

The problem with your argument is that by writing $\{a_{n_k}\}$ and $\{b_{n_k}\}$ you are implicitly (and incorrectly) assuming that the convergent subsequences of $\{a_n\}$ and $\{b_n\}$ involve the same terms. In general, there is no reason why this should be the case. We need to introduce all of the subsequences that we do to get around this problem.
EDIT: Here's a specific example.
Take $a_n$ to be $0$ for $n$ odd, and $(-1)^{n/2}$ for $n$ even. Take $b_n$ to be $(-1)^{(n+1)/2}$ for $n$ odd, and $0$ for $n$ even. These are bounded sequences. Bolzano-Weierstrass guarantees that some subsequence converges (and indeed, there are plenty of convergent subsequences, with different limit points). One possibility would be to take the terms $a_n$ for $n = 0,4,8,16,...$ and $b_n$ for $n = 0,2,4,6,...$ In other words, $a_{n_k} = a_{4k}$, and $b_{n_k} = b_{2k}$. Therefore, your candidate subsequence $\{a_{n_k} b_{n_k}\}$ of $\{a_n b_n\}$ would be the sequence $\{a_{4k} b_{2k}\}$. But this doesn't make any sense, because the terms of the product sequence $\{a_n b_n\}$ are products where the indices on the $a$'s and $b$'s match. 
