Let $(r_n)$ be an enumeration of $\mathbb{Q}$. For each n, let $I_n$ be the open interval cenetered at $r_n$ of radius $2^{-n}$ Let $(r_n)$ be an enumeration of $\mathbb{Q}$. For each n, let $I_n$ be the open interval cenetered at $r_n$ of radius $2^{-n}$ and let $U = \cup_{n=1}^{\infty} I_n$. Prove that $U$ is a proper, open, dense subset of $\mathbb{R}$ and that $U^c$ is nowhere dense in $\mathbb{R}$.
Attempt :
Its a countable union of open sets and hence is open. 
Given a point $x \in \mathbb{R}$ any $\epsilon$ ball containing $x$ would contain a rational number hence $U$ is dense. 
How do you prove $U$ is proper, $U^c$ is nowhere dense ?
There is a partial answer in mathstackexchange which I am not able to understand 
https://math.stackexchange.com/q/10093
 A: The (Lebesgue) measure of $U$ is bounded above by $\mu(U) \le \sum_n \mu(I_n) = \sum_n \frac{2}{2^n} = 2\sum_n \frac{1}{2^n} \le 2$. So $U$ is proper as its measure is not $\infty$.
added on request
If you haven't studied Lebesgue measure yet: for every $n$ pick successive non-empty closed subsets $C_n$ with $C_{n+1} \subset C_n$ such that $\operatorname{diam}(C_n) \le 4r_n$ and $C_n \cap I_n = \emptyset$ this can be done as $\operatorname{diam}(I_n) = 2r_n$, so there is always room for the next $C_{n+1}$ if we have already defined $C_n$ to avoid $I_{n+1}$. Then Cantor's theorem (see the complete metric variant here) about decreasing closed subsets of decreasing-to-$0$ diameters in a complete metric space (proof summary: pick $x_n \in C_n$ show it is Cauchy, and its limit is in) the intersection of the $C_n$ is non-empty and misses $\cup_n I_n$ by construction. As an historical aside: Cantor used this theorem originally to see that $\mathbb{R}$ is uncountable and is his first proof of this fact. Here we avoid an enumeration of open intervals that keep getting smaller and smaller fast, he avoided $r_n$ for a supposed enumeration $r_n$ of $\mathbb{R}$ to get a non-enumerated real...
$U$ is dense as $\mathbb{Q} \subset U$.
The complement of an open and dense set is closed and nowhere dense: Suppose $O \subseteq U^c$,were open and non-empty, then it would have to contain some rational, but these are all in $U$, not in $U^c$. So $\operatorname{int}(U^c) = \emptyset$, which is enough to show nowhere denseness for closed sets. 
Similarly define $U_k$ as $U$ only using $\frac{1}{k}2^{-n}$ as radius around $q_n$, then a similar computation shows that $\mu(U_k) \le \frac{2}{k}$ ,the $U_k$ are still open and dense, and so $D:= \cap_k U_k$ has measure $0$ while $D^c = \cup_k (U_k)^c$ is a countable union of nowhere dense sets, hence is of first category (a.k.a "meagre"). This shows that we can write $\mathbb{R}$ as the union of two "small sets", one small in measure (measure $0$), the other small in "category" (meagre). For more exploration of differences and anlogies between these concepts, see Oxtoby's classical booklet "Measure and category" which starts with this example, in fact.
