Let $\sum_{n=m}^{\infty}a_{n}$ and $\sum_{n=m}^{\infty}b_{n}$ be two formal series of real numbers, and suppose that $|a_{n}|\leq b_{n}$ for all $n\geq m$. Then if $\sum_{n=m}^{\infty}b_{n}$ is convergent, then $\sum_{n=m}^{\infty}a_{n}$ is absolutely convergent, and in fact \begin{align*} \left|\sum_{n=m}^{\infty}a_{n}\right| \leq \sum_{n=m}^{\infty}|a_{n}| \leq \sum_{n=m}^{\infty}b_{n}. \end{align*}
MY ATTEMPT
Since $\sum_{n=m}^{M}b_{n}$ converges, it is a Cauchy sequence. Consequently, for every $\varepsilon > 0$, there is a natural number $N\geq m$ such that \begin{align*} q\geq p\geq N & \Rightarrow \left|\sum_{n=m}^{q}b_{n} - \sum_{n=m}^{p}b_{n}\right| = \left|\sum_{n=p+1}^{q}b_{n}\right|\leq \varepsilon\\\\ & \Rightarrow \left|\sum_{n = p+1}^{q}a_{n}\right| \leq \left|\sum_{n=p+1}^{q}|a_{n}|\right| \leq\left|\sum_{n=p+1}^{q}b_{n}\right|\leq \varepsilon \end{align*}
whence we conclude that $\sum_{n=m}^{M}|a_{n}|$ as well as $\sum_{n=m}^{M}a_{n}$ are Cauchy, thus they converge.
Moreover, due to the triangle inequality and the given assumption, one has that \begin{align*} \left|\sum_{n=m}^{M}a_{n}\right| \leq \sum_{n=m}^{M}|a_{n}| \leq \sum_{n=m}^{M}b_{n} \end{align*}
Given that they all converge, we can take the limit to obtain the desired result, and we are done.
Am I missing any formal step? Please let me know if so.
This is nice, using the fact that the partial sums of $b_n$ convergence, all the series converge since there partial sums are all Cauchy. I don't think you missing any steps, it looks tight. If you're interested here's more for your consideration. The converges of $\sum_{n=m}^\infty |a_n|$ can also follow from what I learned as the monotone convergence theorem (not from Lebesgue integration) which says if a real sequences is monotone and bounded it converges. So, $$A_M := \sum_{n=m}^M |a_n| \le \sum_{n=m}^M b_n \le \sum_{n=m}^\infty b_n <\infty$$ is monotone and bounded, and therefore converges. That $\sum_{n=m}^\infty a_n$ convergence absolutely follows exactly from the definition of absolute convergence, since $\sum_{n=m}^\infty |a_n|$ converges. Triangle inequality together with the order-preserving property of limits does the rest, exactly as you ended it.