Prove that every nonempty open subset $G$ of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping closed rectangles. Question:

Prove that every nonempty open subset $G$ of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping closed rectangles, $G= \bigcup_{k=1}^\infty I_k$.

Attempt:
My idea is to tile $\mathbb{R}^n$ with cubes of side 1. Then discard the cubes that do not intersect with $G$. Then repeat. However, I'm not sure if this is the right approach, since it may only approximate $G$. I also would be unsure of how to formalize such an approach. Alternative ideas would be appreciated.
Here's some additional context since there's quite a bit of discussion on the problem. The problem as originally written in Chapter 2 of Lebesgue Integration on Euclidean Space by Frank Jones reads:

Of course, we need to know what a special rectangle is. This is defined as follows (essentially just a closed rectangle):

 A: The statement is false. A proof by Terence Tao is given in his blog post ‘Covering a non-closed interval by disjoint closed intervals’. To summarize:
Consider the one dimensional case. Suppose $\left\{I_k = \left[a_{k},b_{k}\right]\right\}_{k=1}^{\infty}$ is a collection of disjoint closed intervals such that $\mathbb{R} = \bigcup_{k=1}^{\infty}I_k$. Now consider the set
$$E = \bigcup_{k=1}^{\infty}\left\{a_k,b_k\right\}$$
of endpoints of the intervals. Clearly $E$ is countably infinite. Now $E = \mathbb{R}\setminus \bigcup_{k=1}^{\infty}(a_k,b_k)$, and is thus closed. We claim that $E$ is also perfect. That is $E$ is closed with no isolated points. Let $x \in E$. Then $E$ is either a right end point or a left end point but not both. If it is a right endpoint, then it's easy to see that $x$ is the limit of the left endpoints to the right of $x$. There is a theorem in analysis that says that every perfect subset of a complete metric space is uncountably infinite. Several proofs of this theorem are given here: Proof that a perfect set is uncountable
A: Your idea seems to work, maybe with slight changes.
You should discard cubes that are not contained in $G$, rather than do not intersect $G$. 
Why does it work? Call the cubes $(R_i)_{i=1}^\infty$. Because of the above line, no doubt that $$\bigcup_iR_i\subset G.$$On the other hand, since $G$ is open, every $x\in G$ is contained in a ball which is contained in $G$. As the cubes get small enough, one of them will contain $x$ and be contained in the ball.
A: Here's a somewhat different approach:
For each $x \in G$, there exists $r > 0$ such that $B(x,r) \subset G$.
For each $x$, construct a closed rectangle with rational vertices containing $x$ and contained in $B(x,r)$.
Since there are only a countable number of such rectangles and the union of all the rectangles equals $G$, we can write $G = \cup_{k=1}^\infty I_k$ where each $I_k$ is a rectangle with rational vertices. 
Now define $P_1 = I_1$ and $P_k = I_k \setminus \text{int}(\cup_{j=1}^{k-1} I_j)$ for $k = 2, 3, \cdots$.
Then the $P_k$ are non-overlapping and each $P_k$ can be written as the finite union of non-overlapping special rectangles, so the result follows. 
