The book which I am reading says
One consequence of the fact that $m^{*}(\textbf{Q}) = 0$ is that given any $\varepsilon > 0$, it is possible to cover the rationals $\textbf{Q}$ by a countable number of intervals whose total length is less than $\varepsilon$.
After some thought, I came up with the following solution. Since $|\textbf{Q}| = |\textbf{N}|$, we can count the rationals as we count the the naturals. More precise, for each rational number $q_{i}$ we can consider the interval $I_{i} = (q_{i} - \varepsilon/2^{i+2},q_{i} + \varepsilon/2^{i+2})$ which contains $q_{i}$. Consequently, we have that \begin{align*} \textbf{Q} = \bigcup_{i=1}^{\infty}\{q_{i}\}\subseteq\bigcup_{i=1}^{\infty}\left(q_{i}-\frac{\varepsilon}{2^{i+2}},q_{i} + \frac{\varepsilon}{2^{i+2}}\right) \end{align*}
Since $|I_{i}| = 2^{-i-1}\varepsilon$, the total length of such intervals equals
\begin{align*} \sum_{i=1}^{\infty}|I_{i}| = \sum_{i=1}^{\infty}2^{-i-1}\varepsilon = \frac{\varepsilon}{2} < \varepsilon \end{align*}
Is it correctly thought? Am I missing something? Is there another way to prove the proposed statement?