Open set in $\mathbb{R}^n$ is countable union of rectangles Prove that any open set $U \subset \mathbb{R}^n$ can be represented as countable union of open rectangles with sides parallel to the axes.
Proof: Let $U$ be an open set in $\mathbb{R}^n$ then for any $x=(x_1,x_2,\dots,x_n)\in U$ exists $\varepsilon_x>0$  such that $N_{\varepsilon_x}(x)\subset U$. But rectangle $V^{\varepsilon_x}_x=(x_1-\frac{\varepsilon_x}{\sqrt{n}},x_1+\frac{\varepsilon_x}{\sqrt{n}})\times \dots\times(x_n-\frac{\varepsilon_x}{\sqrt{n}},x_n+\frac{\varepsilon_x}{\sqrt{n}})$ lies in $N_{\varepsilon_x}(x)$.
It's obvious that collection $\{V^{\varepsilon}_x: x\in \mathbb{R}^n, \varepsilon>0\}$ is base of $\mathbb{R}^n$.  Using this problem we conclude that $U$ can be represented as countable union of some subset of this collection.
EDIT: Every open set $U\subset \mathbb{R}^n$ we can write in the following way: $U=\bigcup \limits_{x\in U}V_x^{\varepsilon_x}$
Is my proof correct?
Would be very thankful for comments and remarks!
 A: You can use one extra fact, a so called thinning out lemma: if a space has a countable base $\mathcal{B}$, then any other base $\mathcal{B}'$ has a countable subfamily that is also a base. See this question, which has two valid answers, one of which mine.
The fact that you have shown the rectangles are a base, plus the fact there is some countable base, shows that we can thin out the rectangles to a countable base, and so we can use that countable subfamily to form the unions.
There is also another way, which might be easier (but the thinning out is a useful fact in general): suppose $O = \cup_{i \in I} U_i$ where the $U_i, i \in I$ are open and $I$ is any index set, and suppose we also have some countable base $B_n, n \in \mathbb{N}$. Then for every $x \in O$, pick $i(x) \in I$ such that $x \in U_{i(x)}$, and then pick $n(x) \in \mathbb{N}$ such that $x \in B_{n(x)} \subseteq U_{i(x)}$, as we have a base. 
Having done that, consider $N' = \{n(x) : x \in O \}$ which is countable as a subset of $\mathbb{N}$. For every $n \in N'$, $n = n(x)$ for some $x \in O$, pick one, and define $I'$ to be the set of all $i(x)$ of the $x$ chosen this way.
Then $I'$ is at most countable, as we pick one element for each $n \in N'$. And $O = \cup_{i \in I'} U_i$. Why? right to left inclusion is obvious, as we omit sets, and if $x \in O$, we have $i(x), n(x)$ as before such that $x \in B_{n(x)} \subseteq U_{i(x)}$ and this shows that $n = n(x)$ (for this $x$) is of course in $N'$. So we picked some $x' \in O$ such that $n = n(x')$, and so also $x \in B_{n(x)} = B_{n(x')} \subseteq U_i(x')$, and $i(x') \in I'$ by construction. This proves the other inclusion. 
We have shown that in a second countable space every family of open sets has a countable subfamily with the same union. This property is known as being "hereditarily Lindelöf".  
Note that both of these proofs heavily use choice. So we could also use an ad hoc proof that immediately specialises to rectangles with rational endpoints, using the fact the rationals are dense in the reals. But these general facts are also quite useful to know.
A: It's correct. You just need to work with rational numbers, since they are dense you can approximate any number you want, and therefore you are done.
So you pick rectangles with rational centre and sides of rational length
