Let $\mathcal{J}^n$ by the collection of all "rectangles" in $\mathbb{R}^n$, that is:
$[[a,b))\in\mathcal{J}^n\iff [[a,b))=[a_1,b_1)\times[a_2,b_2)\times\cdots\times[a_n,b_n)$ where $a,b\in\mathbb{R}^n$ and $a_i\le b_i\ \forall i$
I will call such $[[a,b))$ "rectangles".
$R(\mathcal{J}^n)$ denotes the ring generated by the collection of n dimensional rectangles.
A ring is http://www.maths.kisogo.com/index.php?title=Ring_of_sets - simply put, a class of sets closed under set-subtraction and union.
Ring generated by can be found at http://www.maths.kisogo.com/index.php?title=Ring_generated_by along with the proof that given $S\in R(\mathcal{J}^n)$ that there is a finite covering using sets in $\mathcal{J}^n$, that is:
$$S=\bigcup^n_{i=1}[[a_i,b_i))$$
Question:
I need to show that given a finite covering, there is a finite DISJOINT covering. this is obvious, but is difficult to prove.
You can ignore everything below this line, it is just my proof of work
Example
Take the rectangle $[[0,5))\subset\mathbb{R}^2$ that is the "square" $\{(x,y)|0\le x< 5,\ 0\le y< 5\}$
It is easy to see $[[0,5))-[[1,6))=[[0,1))\cup [0,1)\times[1,5)\cup [1,5)\times [0,1)$ for example.
Yet $[[0,5))-[[1,2))$ has yet more cases (8 infact), a cube less something inside of it has 26 chunks.
What I think I must do
I think I need to do something by induction, and consider $\mathcal{J}^n=\mathcal{J}^n\times\mathcal{J}^1$.
What I have done
I have shown this to be true for $\mathcal{J}^1$ (and could do it for any any specific n).
How?
Given a covering $\cup^m_{i=1}B_i$ where $B_i\in\mathcal{J}^1$ we may generate a disjoint covering as follows:
Define $A_1=B_1$ and $A_n=B_n-\cup^{n-1}_{i=1}A_i$ (notice $\cup^n_{i=1}A_i=\cup^n_{i=1}B_i$)
Let us proceed by induction:
$A_1$ can be expressed directly as an interval in $\mathcal{J}^1$
Assume $A_n$ can be expressed as the union of disjoint members of $\mathcal{J}^1$, then
$A_{n+1}=S_{n+1}-\bigcup^n_{i=1}A_i=S_{n+1}\cap[\cup_{i=1}^nA_i]^c$
Then WLOG you can order the (disjoint) intervals present in $\cup_{i=1}^nA_i$, then the complement takes the form $(-\infty,a_1)\cup[b_1,a_2)\cup\cdots\cup[b_{n-1},a_n)\cup[b_n,\infty)$ and as $S_{n+1}=[x,y)$
You don't include all the intervals whos endpoints are below $x$ you cut the one whos lower bound is $\le x$, you keep including while the upper bound is $< y$ then you cut the one which contains $y$, then you ignore the remainder.
A subset of a finite set is finite, so we have (by induction) shown we can cover a finite covering by a finite collection of disjoint members from $\mathcal{J}^1$
Hint
Let $m$ be any positive natural number and let $A_1$, $C_1$, ..., $A_m$, $C_m$ be arbitrary sets. Use the well order to prove that
$(A_1 \times A_2 \times A_3 \times A_4 \times ... \times A_{m-1} \times A_m) \setminus (C_1 \times C_2 \times C_3 \times C_4 \times ... \times C_{m-1} \times C_m) =$
$= [(A_1 \setminus C_1) \times A_2 \times A_3 \times A_4 \times ... \times A_{m-1} \times A_m] \dot{\cup} $
$ \dot{\cup} [(A_1 \cap C_1) \times (A_2 \setminus C_2) \times A_3 \times A_4 \times ... \times A_{m-1} \times A_m] \dot{\cup}$
$\dot{\cup} [(A_1 \cap C_1) \times (A_2 \cap C_2) \times (A_3 \setminus C_3) \times A_4 \times ... \times A_{m-1} \times A_m] \dot{\cup} ... $
$ ... \dot{\cup} [(A_1 \cap C_1) \times (A_2 \cap C_2) \times (A_3 \cap C_3) \times ... \times (A_{m-1} \cap C_{m-1}) \times (A_m \setminus C_m)]$
(disjoint union)
So you have this applied to intervals gives some natural way to write those as disjoint unions of rectangles. Proceed by induction.