Proof by induction of the sum $(a+b+c+d+...)^2$

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$$\text{Assertion:}\;\left(\sum_{i=1}^n |a_i|\right)^2\overset{(*)}{=}\sum_{i=1}^n |a_i|^2+2\sum_{i<j}^n |a_i|\cdot|a_j|$$

whereby $i,j\in\mathbb{N}$ and $a\in\mathbb{R}$. First I'll write what I've accomplished so far:

Let $n=2$ (base clause), then $(\sum_{i=1}^2 |a_i|)^2=(|a_1|+|a_2|)^2=|a_1|^2+|a_2|^2+2|a_1|\cdot|a_2|$. (so the base clause is true).

Now assume that for $n = k$, that $(*)$ holds (induction hypothesis).

$$ \begin{align} n\rightarrow n+1: \left(\sum_{i=1}^{n+1} |a_i|\right)^2 &=\left(\sum_{i=1}^{n+1} |a_i|\right)\left(\sum_{i=1}^{n+1} |a_i|\right)\\ &=\left(|a_{n+1}|+\sum_{i=1}^n |a_i|\right)\left(|a_{n+1}|+\sum_{i=1}^n |a_i|\right)\\ &=|a_{n+1}|^2+2|a_{n+1}|\left(\sum_{i=1}^{n} |a_i|\right)+\left(\sum_{i=1}^{n} |a_i|\right)^2\\ &\overset{(*)}{=}|a_{n+1}|^2+2|a_{n+1}|\left(\sum_{i=1}^{n} |a_i|\right)+\sum_{i=1}^n |a_i|^2+2\sum_{i<j}^n |a_i|\cdot|a_j|\\ &=\sum_{i=1}^{n+1} |a_i|^2+2\left(\sum_{i=1}^n |a_i|\cdot|a_{n+1}|+\sum_{i<j}^n |a_i|\cdot|a_j|\right)\\ \end{align} $$

and that's where I got stuck. Any ideas how to proceed or did I something wrong/need to reconsider?

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You did all the dirty work correctly. Now just notice that the sums in the brackets can be combined into the final form of the sum that you need.

In other words,

$$ \sum_{i=1}^n |a_i|\cdot|a_{n+1}|+\sum_{i<j}^n |a_i|\cdot|a_j| = \sum_{i<j}^{n+1} |a_i|\cdot|a_j| $$