Since $b_n\to 0$ and $b_n>0$ for all $n$, we know that $b_n<\epsilon$ for given $\epsilon >0$.
You've suddenly moved from using $n$ as an unrestricted value ("for all $n$") to using it in ways that are evidently not universal ($b_n<\epsilon$) without properly stating so. Also, the most obvious interpretation (for someone who doesn't know how to do the proof themselves and is therefore trying to learn it from your proof) would be that you mean a fixed value of $n$, dependent only on $\epsilon$.
Since $|a_n−a_m|≤b_n$ for all $m\ge n, |a_n−a_m|<\epsilon$ for all $m\ge n$.
Okay. So now we know that for any $m$ greater than that fixed $n, |a_n−a_m|<\epsilon$.
Let $N\le n$. Then $N\le m$, so for all $m \ge n \ge N, |a_n−a_m|<\epsilon$, which fulfills the definition of a Cauchy sequence.
Really? I can just pick a value $N \le n$ and this works?. Well, I pick $N = 0$. Now we have two ways of interpreting the next sentence:
- For all $m$, but still with that same fixed $n$, when $m \ge n \ge N, |a_n-a_m| < \epsilon$. This is what you showed before. but it does not fulfill the definition of a Cauchy sequence, because that definition requires $n$ to vary as well.
- For all $m$, and for all $n$ with $m\ge n\ge N$. But now, you have no longer shown $|a_m - a_n| < \epsilon$. That was only shown to be true for the fixed value of $n$ chosen earlier. Even worse, since $N$ was arbitrarily chosen, you are claiming it is true even for lower values of $n$ than the one you picked originally. So you have not met the Cauchy condition here, either.
$N$ is the linch-pin of the Cauchy condition. It must be picked with care to satisfy its purpose. It should be picked based on $\epsilon$ right at the start.
Fortunately $b_n \to 0$ means more than $b_n < \epsilon$ for some $n$. The definition of convergence here (using that $b_n > 0$ for all $n$) is
"Given $\epsilon > 0$, there is an $N$ such that for all $n \ge N, b_n < \epsilon$".
Now, $N$ is the fixed value, and $n$ still varies freely over its restricted domain.