How to show this cover of $\mathbb{Q}$ doesn't cover $\mathbb{R}$? Let $\{q_n : n \in \mathbb{N}\}$ be an enumeration of $\mathbb{Q}$ and define $\mathcal{O} = \{I_n : n \in \mathbb{N}\}$ being
$$I_n = \left(q_n - \frac{1}{2^n}, q_n + \frac{1}{2^n}\right).$$
It is obvious that $\mathcal{O}$ is an open cover of $\mathbb{Q}$, but I want to show this is not a cover of $\mathbb{R}$. One possible way that I've already seem is to see that the total length of the intervals is $2\sum_{n=1}^\infty 2^{-n}=2$ while the length of $\mathbb{R}$ is $+\infty$.
Although this works, I'm trying to find another way to prove this result. One thing that is intuitively clear is that if $n\to \infty$ then the intervals shrink as small as desired around the rational midpoint, since $1/2^n \to 0$ as $n\to \infty$. I thought on using this to show that there is some irrational in the "middle" of two such $I_n$ for large enough $n$, but I don't even know how to start this.
Anyway, how can I prove this result without using the measure argument? Is my idea correct? If so, how can it be made rigorous?

Let us work with the closed interval $[0,2]$, let $\{q_n\}$ be a enumeration of $\mathbb{Q}\cap [0,2]$, and define 
$$I_n := \bigg(q_n -\frac{1}{2^{n+1}}, q_n+ \frac{1}{2^{n+1}}\bigg)\bigcap [0,2]$$
then $\{I_n\}$ is an open cover of the rationals in the space $[0,2]$.
(I used  $\frac{1}{2^{n+1}}$ so that the sum of the length is $1$, there is a small typo in your post.)
Since $I_n^c$ the complement in $[0,2]$ is closed, define 
$$C_k =\bigg( \bigcup_{n=1}^k I_n\bigg)^c = \bigcap_{n=1}^k I_n^c,$$
we see that each $C_k$ is non-empty, closed, and the sequence $\{C_k\}$ is nested. Therefore, using nested set theorem (Cantor's intersection theorem)
$$\text{ there exists } x\in \bigcap_{n=1}^\infty I_n^c \neq \emptyset$$
by construction, $x$ is irrational, and therefore $\{I_n\}$ doesn't cover all irrational numbers.
For the argument in $\mathbb{R}$, you can first restrict to the compact subspace $[0,2]$, where your open cover of $\mathbb{Q}$ would become $I_n\cap [0,2]$ and the same argument would follow. 
