$$|a_jb_j-a_kb_k|=|b_j(a_j-a_k)+a_k(b_j-b_k)|\le |b_j||a_j-a_k|+|a_k||b_j-b_k|\le A\delta+B\delta'$$ where $A,B$ are upper bounds on $|a_j|$ and $|b_k|$ for $j>M$ and $k>M'$ (say $|a_M|+\delta$ and $|b_{M'}|+\delta'$).
To achieve a given $\delta''$, you can take $\delta=\dfrac{\delta''}{2A},\delta'=\dfrac{\delta''}{2B}$.
I don't have enough time to study your argument in detail; I hope Joshua P. Swanson's answer addressed your problems here. However, I would recommend adding some structure to your argument by dividing the proof in smaller pieces.
For example, are you familiar with the real case of your problem? If yes, the complex case is easily reduced (see Step 2). If not, see Step 1 and try to tackle it separately, without the notational noise coming from the more general case.
Step 1. If $\sum_{k=1}^\infty x_k$ is a real conditionally convergent series, then after rearranging $\sum_{k=1}^\infty x_{\varphi(k)} = +\infty$.
This is a special case of the Riemann rearrangement theorem. The construction is the following: first start the rearrangement with as many positive $x_k$'s as is needed to obtain a sum larger than $1$. Second, add one nonpositive $x_k$. Next, add some positive terms to obtain a sum larger than $2$, add one nonpositive term, and so on.
Step 2. The complex case.
Denote the real and imaginary part: $z_k = a_k + i b_k$. Since $\sum z_k$ converges, both $\sum a_k$ and $\sum b_k$ converge as well. If also both $\sum |a_k|$ and $\sum |b_k|$ converged, the inequality $|z_k| \le |a_k|+|b_k|$ would imply that $\sum |z_k|$ convergences too - and we know this is false. So we may assume (without loss of generality) that $\sum a_k$ converges only conditionally.
Taking the rearrangement as in Step 1 (for $x_k = a_k$), we obtain
$$
\left|\sum_{k=1}^N z_k\right|
= \left|\sum_{k=1}^N a_k + i \sum_{k=1}^N b_k\right|
\ge \left|\sum_{k=1}^N a_k\right| \xrightarrow{N \to \infty} \infty,
$$
thanks to the inequality $|a+ib| \ge |a|$.
Best Answer
The right-hand side of the first equality is \begin{align*} (RHS)&=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\sum_i\sum_j(a_ib_j-b_ia_j)^2\\ &=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\sum_i\sum_j(a_i^2b_j^2-2a_ib_ia_jb_j+b_i^2a_j^2)\\ &=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\sum_i\sum_ja_i^2b_j^2+\sum_i\sum_ja_ib_ia_jb_j-\frac{1}{2}\sum_i\sum_jb_i^2a_j^2\\ &=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\left(\sum_ia_i^2\right)\left(\sum_jb_j^2\right)+\left(\sum_i a_ib_i\right)\left(\sum_ja_jb_j\right)-\frac{1}{2}\left(\sum_ib_i^2\right)\left(\sum_ja_j^2\right) \end{align*} Change all summation indices to $k$: \begin{align*} (RHS)&=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)+\left(\sum_k a_kb_k\right)\left(\sum_ka_kb_k\right)-\frac{1}{2}\left(\sum_kb_k^2\right)\left(\sum_ka_k^2\right)\\ &=\left(\sum_ka_kb_k\right)^2 \end{align*}