TITLE: Proof Verification About Sequence Limits
QUESTION [7 upvotes]: I had work to hand in which included the following question

Let $(a_n), (b_n)$ satisfy $\lim_{n\rightarrow\infty} a_nb_n=1$.
Proof that if for all $n$, $0\leq a_n,b_n \leq 1$ then
   $\lim_{n\rightarrow\infty} a_n=\lim_{n\rightarrow\infty}b_n=1$.

My solution which I handed in was pretty straightfoward- showing easily that $a_nb_n\leq a_n\leq 1$ and the same for $b_n$. Then I used the squeeze theorem to show their limits are $1$.
However, people (also the teachers who assisted some students) started talking about partial limits and Bolzano–Weierstrass which really makes my dbout my solution's validness, which I seek here from you at the moment.
Thanks in advance.
EDIT See my comment below

REPLY [4 votes]: Your argument is fine and is the easiest one that occurs to me.
An argument using the Bolzano-Weierstrass theorem is also possible, though it takes a bit more work. Suppose that $\langle a_n:n\in\Bbb N\rangle$ does not converge to $1$. Then there is an $\epsilon>0$ such that $M=\{n\in\Bbb N:a_n\le 1-\epsilon\}$ is infinite. The subsequence $\langle a_n:n\in M\rangle$ is bounded, so by the Bolzano-Weierstrass theorem it has a convergent subsequence: there is an infinite $M_0\subseteq M$ such that $\langle a_n:n\in M_0\rangle$ converges, say to $\ell$; clearly $\ell\le 1-\epsilon$, since $a_n\le 1-\epsilon$ for each $n\in M_0$. But $\langle a_nb_n:n\in M_0\rangle$ converges to $1$, so $\langle b_n:n\in M_0\rangle$ converges to $\frac{1}{\ell}>1$, which is impossible, since $b_n\le 1$ for each $n\in\Bbb N$. This contradiction shows that $\lim_na_n=1$, and it follows immediately that $\lim_nb_n=1$ as well.