In order to show that $\sum_{i=1}^\infty\sum_{j=1}^\infty|a_ib_j|$ converges, you must show that there is some real number $a$ such that for every $\epsilon>0$ there is some $n_0\in\Bbb Z^+$ such that $\left|a-\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\right|<\epsilon$ whenever $m,n\ge n_0$.
For $m,n\in\Bbb Z^+$ let $s_{mn}=\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\;.$ You know that the set $S=\{s_{mn}:m,n\in\Bbb Z^+\}$ is bounded. You also know that $s_{k\ell}\le s_{mn}$ whenever $k\le m$ and $\ell\le n$, since the terms $|a_ib_j|$ are non-negative. For $n\in\Bbb Z^+$ let $d_n=s_{nn}$; then $s_{k\ell}\le d_n$ whenever $k,\ell\le n$. In other words, the sums $d_n$ are cofinal in $S$: for each $s\in S$ there is an $n\in\Bbb Z^+$ such that $s\le d_n$.
Consider the sequence $\langle d_n:n\in\Bbb Z^+\rangle$: it’s bounded and non-decreasing, so it converges to some limit $a$; I’ll leave to you the easy verification that $s\le a$ for all $s\in S$. Fix $\epsilon>0$. There is some $n_0\in\Bbb Z^+$ such that $|a-d_n|<\epsilon$ whenever $n\ge n_0$. Suppose that $m,n\ge n_0$; then $a-\epsilon, so $\left|a-\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\right|<\epsilon$ whenever $m,n\ge n_0$, exactly as we wanted.