TITLE: A classic example in measure theory, hold the measure theory
QUESTION [8 upvotes]: I answered a question recently drawing on the following construction. Let $\{r_n\}$ be a enumeration of the rationals and set
$$E:=\bigcup_{n=1}^\infty B(r_n,2^{-n}).$$
Then 
$$m(E) \leq \sum_{n=1}^\infty m\big(B(r_n,2^{-n})\big)=2$$
and in particular $\mathbb R \setminus E$ is uncountable. I realized when answering this question that I didn't know how to prove that $\mathbb R \setminus E$ is uncountable without measure theory. It seems to me a nice example in point set topology, I recall a period of time when I thought that surely any dense open subset of $\mathbb R$ was only missing a countable number of points (I hadn't read about the Cantor set either at that time). So I was wondering if there was a way to approach this that a student in a typical first class in analysis or point-set topology would understand.

REPLY [4 votes]: It is not too hard to prove that if $( (a_i,b_i) : i \in I )$ is an open cover of the unit interval $[0,1]$ then $\sum_{i \in I} (b_i - a_i) > 1$.  The proof is just a compactness argument: there would be some finite subcover $( (a_j, b_j) : j < k )$ of the interval, and by considering the endpoints we can show that $\sum_{j < k} (b_j - a_j)$ is already greater than $1$. 
This is of course half of the proof that $m([0,1]) = 1$, but you could present it without defining measure at all.