I would be very grateful if someone would verify whether my proof below is correct.
Many thanks.
Theorem. $\,$ Let $(b_k)$ be a rearrangement of the complex sequence $(a_k)$. If $\sum_{k\geq 0}a_k = s$ and is absolutely convergent, then $\sum_{k\geq 0}b_k = s$.
Proof. $\,$ Let $\varepsilon>0$ be given. Choose $n\geq 0$ such that
$$\sum_{k=n+1}^\infty|a_k|<\varepsilon.$$
Choose $N\geq 0$ such that
$$\{a_1,\ldots,a_n\}\subseteq\{b_1,\ldots,b_N\},$$
Then, for any $m\geq N$, we have that
$$\sum_{k=0}^{\infty}a_k-\sum_{k=0}^mb_k=\sum_{k\in A_m}a_k.$$
where $A_m=\{n+1,n+2,\ldots\}\setminus\{\text{finitely many points}\}$.
Hence, for any $m\geq N$, it follows that
$$\left|\sum_{k=0}^{\infty}a_k-\sum_{k=0}^mb_k\right|\leq\sum_{k=n+1}^\infty|a_k|<\varepsilon.$$
This is what we were required to prove.
Yes, the proof you've provided is correct.