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?
 A: 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. 
A: This answer is kind of cheating, since it's really just extracting the essential pieces of measure theory needed to prove this.  But it's self contained and doesn't need to use the word "measure".
Fix a closed interval such as $[0,2]$ whose length is at least as long as the sum of the lengths of your intervals $I_n$ (as currently formulated, that sum is 2 instead of 1). By compactness (as suggested in Xiao's answer), it is sufficient to show that no finite subset of $\mathcal{O}$ covers $[0,2]$.  This can be done by means of the following elementary lemma:
Lemma. For every $n$, every interval $[a,b]$, and every set of $n$ intervals $(a_1, b_1), \dots, (a_n, b_n)$  such that $[a,b] \subset \bigcup_{i=1}^n (a_i, b_i)$, we have $\sum_{i=1}^n (b_i-a_i)>b-a$.
Proof.  Proceed by induction on $n$.  The base case $n=1$ is easy.  Now suppose the claim holds for $n-1$.  Let $[a,b]$ be an interval covered by $(a_1, b_1), \dots, (a_n, b_n)$.  One of the intervals $(a_i, b_i)$ must contain the point $b$; without loss of generality suppose it's $(a_n, b_n)$.  If $a_n < a$ we are done, since then $(a_n, b_n)$ covers $[a,b]$ and we invoke the $n=1$ case to see that $b-a < b_n-a_n \le \sum_{i=1}^n (b_i-a_i)$.  Otherwise, if $a_n \ge a$, then the intervals $(a_1, b_1), \dots, (a_{n-1}, b_{n-1})$ cover $[a, a_n]$.  So by the inductive hypothesis, $\sum_{i=1}^{n-1} (b_i - a_i) > a_n - a$.  Now adding $b_n - a_n$ to both sides and noting that $b_n > b$ we have
$$\sum_{i=1}^{n} (b_i - a_i) > (a_n - a) + (b_n - a_n) = b_n - a > b - a$$
as desired.
(This is essentially the proof that shows that the Lebesgue outer measure of an interval $[a,b]$ is $b-a$, which is a key step in the construction of Lebesgue measure.)
Let me also put in an advertisement for a pretty approach by David C. Ullrich in Lebesgue Premeasure via Transfinite Induction, which uses completeness instead of compactness and so dodges the Heine-Borel theorem.
