Let $B$ be the set of rational numbers in the interval $[0,1]$, and let $\{I_k\}_{k=1}^n$ be a finite collection of open intervals that covers $B$. Prove that $\sum_{k=1}^n m^*(I_k) \geq 1$.
Proof:
First, we have $B = \mathbb{Q}\cap [0,1] \subset [0,1]$ and $B \subset \bigcup_{k=1}^n I_k$, where each $I_k$ is an open interval. It follows immediately that since $B$ is countable has outer measure zero and
$$B \subset [0,1] \Rightarrow m^*(B) \leq m^*([0,1]) = 1,$$
$$B \subset \bigcup_{k=1}^n I_k \Rightarrow m^*(B) \leq m^*(\bigcup_{k=1}^n I_k) \leq \sum_{k=1}^{\infty} m^*(I_k).$$
Now, since $0 \in B$, there exist one of the $I_k$'s that contains $0$, denote such interval by $(a_1,b_1)$. By a similar reasoning, there exist one of the $I_k$'s that contains $1$ (Note that $1 \in B$), denote such interval by $(a_N,b_N)$. We thus have:
$$a_1 < 0 < b_1$$
$$a_N < 1 < b_N$$
Now, lets assume the contrary, that is $\sum_{k=1}^n m^*(I_k) < 1$. In particular, we are assuming the following:
$$\sum_{k=1}^n m^*(I_k) < m^*([0,1])$$
We claim that if $b_1 \geq 1$, we obtain a contradiction, since:
$$a_1 < 0 < 1 \leq b_1$$
and
$$ l((a_1,b_1)) = b_1 - a_1 \leq \sum_{k=1}^n m^*(I_k) < 1,$$
but $b_1 - a_1 > 1$. Otherwise, $b_1 \in [0,1)$ and since $b_1 \notin (a_1,b_1)$, there exist an interval in the collection $\{I_k\}_{k=1}^{n}$ which we label as $(a_2,b_2) \ni b_1$ and $$a_2 < b_1 < b_2.$$
We claim that if $b_2 \geq 1$, we obtain a contradiction, since:
$$b_2 - a_1 < (b_2 -a_2)+(b_1 -a_1) \leq \sum_{k=1}^n m^*(I_k) < 1$$
but $b_1 -a_0 > 1$ because $$a_0 < 0 < 1 \leq b_1.$$
We can continue this selection process until it terminates, as it must since there are only $n$ intervals in the collection $\{I_k\}_{k=1}^{n}$. We thus obtain a sub-collection $\{(a_k,b_k)\}_{k=1}^N$ of $\{I_k\}_{k=1}^{n}$ for which $a_1 < 0,$ while $a_{k+1} < b_k$ for $1 \leq k \leq N-1$ and $b_N>1$. We thus have:
$$\begin{align*}
b_N-a_1 &< b_N - (a_N-b_{N-1})- \cdots-(a_2-b_1) -a_1 \\
&\leq (b_N-a_N)+(b_{N-1}-a_{N-1})+\cdots +(b_1-a_1) \\
&= \sum_{k=1}^N l(I_k) \leq \sum_{k=1}^n m^*(I_k)<1
\end{align*}$$
a contradiction, since $$l((0,1)) = 1 < b_N-a_1$$ and $$a_1 < 0 < 1 < b_N.$$
Thus, we conclude that: $$\sum_{k=1}^n m^*(I_k) \geq 1.$$
The argument in your first paragraph is correct.
For the second paragraph, why do you think that being disconnected is of any relevance? The rational numbers are of zero measure because they are countably many of them. The set of irrationals is not countable, therefore it can (and indeed does) have a non-zero measure.
On your third paragraph: It is true that between any two rationals there's an irrational, and between any two irrational there's a rational. However, there is not only one irrational between two rationals, and not only one rational between two irrationals. Rather there are countably many rationals between two irrationals, but uncountably many irrationals between two rationals.
In other words, it's not that you have an alternating sequence of rational and irrational numbers (which is what you seem to have in mind). In particular, you cannot define ”the next larger rational” nor ”the next larger irrational.”
In some sense, although the rationals are already dense, the irrationals are in a sense even more dense.
Maybe it helps if instead of the rationals, you consider the numbers with finite decimal expansion, which also is a countable dense subset. Here, it might be more intuitive that, you have many more possibilities for numbers if you can arbitrarily continue your number string infinitely than if you are forced to stop after finitely many digits. Yet again, between any two finite-expansion numbers, you always can find an infinite-expansion number (just add infinitely many non-zero digits to the smaller one, after first padding it to the length of the larger with zeroes if necessary), and between any two infinite-expansion numbers you can find a finite-expansion number (just cut the larger one at an appropriate point).
Best Answer
You can do even better, since $A=\mathbb{Q}\cap (0,1)$ has measure $0$, so $\forall \epsilon>0$, we can find intervals $I_n$ such that $A\subset \bigcup I_n$ and $\sum l(I_n)\leq\epsilon$ so that $m(A)\leq \epsilon$. The idea is that $A$ is countable, so $A=\{q_n\}_{n\in\mathbb{N}}$. Now let $$I_n=\left(q_n-\frac{\epsilon}{2^{n+1}},q_n+\frac{\epsilon}{2^{n+1}}\right)$$ It is easy to see that $\sum l(I_n)\leq\epsilon\sum 1/2^{n}=\epsilon$