For $\epsilon > 0$, $\mu((q_n - \epsilon, q_n + \epsilon)) = 2\epsilon$ and $\sum_{n=1}^\infty 2\epsilon = \infty$, so this reasoning doesn't work.
A simple method to show that $\mu(\mathbb Q) = 0$ is to notice that the measure of a single point is $0$, thus:
$$
\mu(\mathbb Q) = \mu\left(\bigcup_{n=1}^\infty \{q_n\}\right) = \sum_{n=1}^\infty \mu(\{q_n\}) = \sum_{n=1}^\infty 0 = 0
$$
The fault in your reasoning is where you claim, without justification, "So $a$ must be in some . . ." Unless you come up with some reason for that "so", there is nothing more to be explained here.
Your conclusion is wrong, those intervals do not cover $\mathbb R.$ That is clear from measure theory, as you know. In order to give an explicit example of a real number which is not covered, we have to know how the rationals are enumerated. Since you did not assume anything special about the enumeration, I can use any enumeration I like. Let me define a special enumeration which makes it easy to exhibit an uncovered number. (Also I will take $\epsilon=1.$)
First, let $q_1,q_2,q_3,\dots$ be your favorite enumeration of the rationals. Now I define a new enumeration $\{r_i\}$ recursively. If $r_1,\dots,r_{i-1}$ have already been defined, then I define $r_i$ to be the first $q_j$ such that $q_j\notin\{r_1,\dots,r_{i-1}\}$ and $|q_j-\sqrt2|\gt\frac1{2^{i+1}}.$
Does the sequence $r_1,r_2,r_3,\dots$ contain all the rationals?
Since your sequence $q_1,q_2,q_3,\dots$ contains all the rationals, it will be enough to show that each $q_j$ occurs in the sequence $r_1,r_2,r_3,\dots.\ $ I will prove this by induction on $j.$ Suppose that each of the numbers $q_1,q_2,\dots,q_{j-1}$ occurs in the sequence $\{r_i\}.$ Since $q_j$ is rational while $\sqrt2$ is irrational, we know that $|q_j-\sqrt2|\gt0.$ Thus (with $j$ fixed) if we choose $i$ large enough, we will have both $\{q_1,\dots,q_{j-1}\}\subseteq\{r_1,\dots,r_{i-1}\}$ and $\frac1{2^{i+1}}\lt|q_j-\sqrt2|.$ For such an $i$ we will certainly have $q_j\in\{r_1,\dots,r_{i-1},r_i\}.$
Do the intervals $(r_i-\frac1{2^{i+1}},\ r_i+\frac1{2^{i+1}})$ cover $\mathbb R$?
No, none of those intervals covers $\sqrt2,$ because $|r_i-\sqrt2|\gt\frac1{2^{i+1}}.$
Best Answer
No, your suggestion is not a probability measure as it is not additive over countable unions of disjoint sets.
If $t$ is a natural number then $m(\{q_t\})= \lim_{n \to \infty} \frac1n=0$ and so $\sum_t m(\{q_t\})=0$. But $m( \mathbb{Q})= \lim_{n \to \infty}\frac{n}{n}=1$.
You need a discrete measure.
For example $\mathbb{Q} \cap (0,1)$ you could use $m \left( \left\{ \frac{a}{b} \right\}\right) =\frac{\zeta(k)}{\zeta(k-1) - \zeta(k) } \left(\frac{1}{b}\right)^k$ for some $k\gt2$ and $a$ and $b$ coprime; this can be extended to all rationals.