Proof

By hypothesis, $\sum_{k=0}^\infty x_k$ is a complex series, $\sum_{k=0}^\infty y_k$ is a convergent real series, and $|x_k|\le y_k$ for all $k\in\mathbb N.$
Since \sum_{k=0}^\infty y_k is a convergent real series, thus we can apply Cauchy's general criterion for the convergence of infinite complex series, and conclude that for every $\epsilon > 0$ there is an index $N(\epsilon)\in\mathbb N$ such that $$\left| \sum_{k=m}^n y_k \right| < \epsilon\quad\quad \text{for all}\quad n\ge m\ge N(\epsilon).$$
Because by hypothesis $|x_k|\le y_k$ for all $k$, we get $$\left|\sum_{k=m}^n |x_k|\right| \le \left|\sum_{k=m}^n y_k\right| < \epsilon\quad\quad \text{for all}\quad n\ge m\ge N(\epsilon).$$
Applying Cauchy's general criterion for the convergence of infinite complex series once again, we get that the $\sum_{k=0}^\infty |x_k|$ is convergent.
Therefore $\sum_{k=0}^\infty x_k$ is an absolutely convergent complex series.
∎

Thank you to the contributors under CC BY-SA 4.0!

Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983