The theorem states that if partial sums $A_n$ of series $\sum a_n$ are bounded and $(b_n)$ is a monotonically decreasing sequence such that $b_n\to 0$, then the series $\sum a_nb_n$ converges.
I tried to prove it as follows:
Since $A_n$'s are bounded, there exists $M\gt 0$ such that $|A_n|\lt M$ for all $n\in \mathbb N$. Since $(b_n)$ converges, for every $\epsilon\gt 0, \exists N:\forall q\ge p\ge N,$ we have $b_p-b_q\lt \frac \epsilon {2M}$.
We have, $$|\sum_{n=p}^q a_nb_n|=|\sum_{n=p}^{q-1}A_n(b_n-b_{n+1})+A_qb_q-A_{p-1}b_p|$$ $$\lt M |\sum_{n=p}^{q-1}(b_n-b_{n+1})+b_p-b_q|=2M(b_p-b_q)\lt \epsilon \tag 1$$
Hence, by Cauchy Criterion, the series $\sum a_nb_n$ converges.
Please note that in the above proof, I have nowhere used $b_n\to 0$. I have only used the fact that the sequence $(b_n)$ is monotonically decreasing and convergent.
My question is why $b_n$ must converge to $0$? Clearly, from the proof above, convergence of $(b_n)$ is enough, irrespective of to what $\alpha\in \mathbb R$, $(b_n)$ converges.
Edit: Suppose that $(b_n)$ is a sequence of non-negative nos. We have $-M\lt A_q\lt M\implies -Mb_q\lt A_qb_q\lt Mb_q$. Similarly, $-Mb_p\lt A_{p-1}b_p\lt Mb_p$. It follows that $-M(b_p-b_q)\lt A_{p-1}b_p-A_qb_q\lt M(b_p-b_q)\implies |A_{p-1}b_p-A_qb_q|\lt M(b_p-b_q)$ so $(1)$ holds. Is it necessary for $b_n\to 0$?
Your edit clarifies the issue in your proof; indeed, if that proof were valid, why, any bounded set of numbers would have to be equal to one another! (set $b_p = b_q = 1$)
The issue is how you are subtracting the inequalities: If $a_l < a < a_u$ and $b_l < b < b_u$, then $a_l - b_u < a - b < a_u - b_l$ and not $a_l - b_l < a - b < a_l - b_l$. Intuitively, the largest $a - b$ can be is if $a$ is its largest, and $b$ is its smallest value.