A closed nonempty subset of $\mathbb{R}$ that contains only irrationals, with complement of length $1$

I found the following question interesting:

Let $\{r_1,r_2,r_3,...\}$ be an enumeration of the rational numbers, and for each $n\in\mathbb{N}$, set $\epsilon_n=1/2^{n}$. Define $O=\bigcup_{n=1}^{\infty}V_{\epsilon_{n}}(r_n)$, and let $F=O^c$.

Argue that $F$ is a closed, nonempty set consisting only of irrational numbers.

Proof. By De Morgan's law, $F=\bigcap_{n=1}^{\infty}V_{\epsilon_{n}}^{c}(r_n)$. Since $V_{\epsilon_{n}}(r_n)$ is an open interval, it follows that $V_{\epsilon_{n}}^{c}(r_n)$ is closed, and since the intersection of an arbitrary collection of closed intervals is closed, it follows that $F$ is closed. Moreover, $O$ has length of at most $1+1/2+1/2^2+...=2$, which implies that $F$ has infinite length and is thus nonempty.

I am clueless as to how to show that $F$ consists only of irrational numbers. What do you guys think?

-
By definition, $O$ contains every rational number. Each $V_{\epsilon_n}(r_n)$ contains $r_n$. Every number on your initial list is in $O$. –  David Mitra Feb 16 '12 at 21:26
The concept of length that you use to prove non-empty is intuitively reasonable, perhaps a little vague. It can be replaced by the more precise concept of measure. Or else (but that would not answer the question as posed) you could modify the definition of the $V(r_n)$ to exclude a specific irrational. –  André Nicolas Feb 16 '12 at 21:50
You can prove $F$ is closed more directly: each $V_{\epsilon_n}(r_n)$ is open (an open ball presumably), so $O$ is open. Thus $F=O^c$ is closed. And, what do you mean by "with complement of length 1" in the title? Did you mean at most 1? Note that, if this is the case, when you were summing the lengths of the $V$'s, you should have started with $1/2$, not $1$ (so $r_1$ is the center of an open ball of radius $1/2$). –  David Mitra Feb 16 '12 at 21:58