I need to find two sequences $(a_n)_{n\geq1}$ and $(b_n)_{n\geq1}$ contained in the interval $[0,1]$ such that $$ \mathbb{Q}\cap (0,1) \subseteq \bigcup_{n\geq 1} (a_n,b_n) \tag{1} $$ and $$ \sum_{n\geq 1} (b_n-a_n) < 1 \tag{2} $$
I think I can find such sequences that satisfy the first condition, for example $\left(\frac{1}{2^n}\right)_{n\geq1}$ and $\left(\frac{n-1}{n}\right)_{n\geq1}$, but I draw a complete blank on how I can also satisfy the second.
$\mathbb{Q}$ is countable, and thus so is $\mathbb{Q}\cap (0,1)$. Let $(r_n)_{n\in\mathbb{N}}$ be an enumeration of the latter.
You can then set, for $n\in\mathbb{N}$, $$ a'_n \stackrel{\rm def}{=} r_n - \frac{1}{3\cdot 2^{n+1}}, \qquad b'_n \stackrel{\rm def}{=} r_n + \frac{1}{3\cdot2^{n+1}} $$ and $$ a_n \stackrel{\rm def}{=} \max(a'_n,0),\qquad b_n \stackrel{\rm def}{=} \min(b'_n,1) $$ (that last step to ensure $a_n,b_n\in[0,1]$) to get what you want.
Let $\{ q_n\}_{n=1}^\infty$ be an enumeration of the rationals in $[0,1]$ and let $0<\epsilon <1 $. take $a_n = q_n - \epsilon/2^{n+2}$, $b_n =q_n + \epsilon/2^{n+2}$. Then $$\sum_{n=1}^\infty b_n - a_n = \sum_{n=1}^\infty \frac{\epsilon}{2^{n+1}} = \epsilon.$$
HINT; Enumerate the numbers from the first set. Then cover the first one by the interval of length $<1/4$, the second — of the length $<1/8$, then $<1/16$ and so on.