Let $\sum_{n=0}^{\infty}{x_{n}}$ be a series in $\mathbb{C}$ that is conditionally convergent (i.e. convergent, but not absolutely). Show that there exists a bijection $\pi : \mathbb{N} \rightarrow \mathbb{N}$ so that the series $\sum_{n=0}^{\infty}{x_{ \pi (n)}}$ does not converge.
We are given these two facts:
Let $(x_{n})_{n}$ be a sequence in $\mathbb{R}$ and let $x_{n}^{+}= \max \{ 0,x_{n} \}$ and $x_{n}^{-}= \max \{ 0,-x_{n} \}$. Then the series $\sum_{n=0}^{\infty}{x_{n}}$ absolutely converges if and only if the series $\sum_{n=0}^{\infty}{x_{n}^{+}}$ and $\sum_{n=0}^{\infty}{x_{n}^{-}}$ both converge.
Let $(x_{n})_{n}$ be a sequence in $\mathbb{C}$ and let $a_{n}=\Re(x_{n})$ and $b_{n}=\Im(x_{n})$. Then the series $\sum_{n=0}^{\infty}{x_{n}}$ absolutely converges if and only if the series $\sum_{n=0}^{\infty}{a_{n}}$ and $\sum_{n=0}^{\infty}{b_{n}}$ both converge absolutely.
Is the following reasoning correct? We know that the new series $\sum_{n=0}^{\infty}{x_{ \pi (n)}}$ does not converge in $\mathbb{C}$. This implies that both $\sum_{n=0}^{\infty}{a_{n}}$ and $\sum_{n=0}^{\infty}{b_{n}}$ do not converge either. It then follows that $\sum_{n=0}^{\infty}{a_{n}^{+}}$, $\sum_{n=0}^{\infty}{a_{n}^{-}}$, $\sum_{n=0}^{\infty}{b_{n}^{+}}$and $\sum_{n=0}^{\infty}{b_{n}^{-}}$ are not convergent.
If this was correct so far, then how can I go on from this? Or is there a “better” solution to this problem?
Your reasoning is not quite correct. The fact that $\sum x_{\pi(n)}$ does not converge means that at least one of the two series $$ \sum a_{\pi(n)} \qquad \mbox{ and }\qquad \sum b_{\pi(n)} $$ diverges (but not necessarily both of them diverge).
A simple proof might
(1) show that conditional convergence of a series $\sum x_n$ in ${\mathbb C}$ implies conditional convergence of at least one of the two series $\sum {\rm Re\,}x_n$, $\sum {\rm Im\,}x_n$ in ${\mathbb R}$ - and then
(2) appeal to Riemann's theorem on conditionally convergent series.