If $(a_n:n\in\mathbb{N})$ is bounded, then for each $N\in\mathbb{N}$, $sup\{a_n:n\geq N\}-inf\{a_n:n\geq N\}\leq sup(|a_n-a_m|\mid m,n\geq N\}$
So I figured that I should use the reverse triangle inequality theorem, i.e. $|a_n-a_m|\geq|a_n|-|a_m|$
Soo for each $n,m\in\mathbb{N}$, we have $sup(|a_n-a_m|)\geq sup(|a_n|-|a_m|)\geq sup(a_n)-sup(a_m)$, but then $sup(a_n)-sup(a_m)\leq sup(a_n)-inf(a_m)$.
What am I missing here?
You went one step further than you want. $$\sup_{m,n\geq N}(\lvert a_n-a_m\rvert)\geq \sup_{m,n\geq N}(a_n-a_m) = \sup_{m,n\geq N}(a_n)+\sup_{m,n\geq N}(-a_m) = \sup_{n\geq N}(a_n)-\inf_{m\geq N}(a_m)$$ The second-to-last step is due to the independence and boundedness of $a_m$ and $a_n$, and the last step is due to this.