Union of boxes in $\bar{\mathbb{R}}^3$ Let $\bar{\mathbb{R}}$ denote the extended real line. Let    $[x_1,x_2]\times[y_1,y_2]\times [z_1,z_2]$ represent a generic "box" in  $\bar{\mathbb{R}}^3$:
$$[x_1,x_2]\times[y_1,y_2]\times [z_1,z_2]\equiv \{(x,y,z) \in \mathbb{R}^3: x_1\leq x \leq x_2,  y_1\leq y \leq y_2, z_1\leq z \leq z_2\}$$
The square brackets can also be replaced with round brackets, depending on whether the extreme points of each interval are included.
Let $\mathcal{A}_1$ be the collection of boxes of the form
$$
(a+b, \infty]\times [-\infty, a]\times [-\infty,b]
$$
for every $(a,b)\in \mathbb{R}^2$.
Let $\mathcal{A}_2$ be the collection of boxes of the form
$$
[-\infty,a+b)\times [a,\infty]\times [b,\infty]
$$
for every $(a,b)\in \mathbb{R}^2$.
Let $\mathcal{A}\equiv \mathcal{A}_1\cup \mathcal{A}_2$.
Consider a box $B$  such that:
(1) $B\notin \mathcal{A}$.
(2) $B$ is not contained in (is not a "sub-box" of) any element of $\mathcal{A}$.
An example of $B$ satisfying 1-2 is
$$
(0.2, \infty]\times [-\infty, 10]\times [-\infty,11]
$$
Or also
$$
[0.2, \infty]\times [-\infty, 10]\times [-\infty,11]
$$
Question: can a box $B$ satisfying 1-2 be contained in (be a sub-box of) the union of some elements of $\mathcal{A}$? The union should not necessarily be a finite union of elements of  $\mathcal{A}$.
Conjecture: I think the answer to my question is no. I don't know how to prove it, though.
Also, I think that the following holds:
$$
\bar{\mathbb{R}}^3\setminus \{\cup_{D\in \mathcal{A}}D\} = \{(x,y,z)\in \mathbb{R}^3: x-y-z=0\}
$$
That is: once we remove from $\bar{\mathbb{R}}^3$ the union of the boxes in $\mathcal{A}$, we get a plane. I'm not sure  how to use this information, though.
 A: If you allow arbitrary combinations of open and closed brackets for $B$, but still restrict the boxes in $\mathcal{A}$ to have open and closed brackets as stated above, then your conjecture is false - a counter-example exists due to a technicality.
Claim: $B = [0, 1] \times [-1, 0) \times [-1, 0)$ satisfy 1-2.
$B \notin \mathcal{A}$ is obvious.
Next consider whether $B$ can be a sub-box of some $A \in \mathcal{A}_1$.  Because the $x$-range of $B$ is $[0,1]$, we need $a + b < 0$ since any box $A$ has $x$-range $(a+b, +\infty]$.  However this implies $a < 0$ or $b < 0$, which would violate either $B$'s $y$-range or $z$-range.
Finally consider whether $B$ can be a sub-box of some $A \in \mathcal{A}_2$.  Similarly $B$'s $x$-range requires $a+b > 1$, which implies $a > 0$ or $b> 0$, which violate either $B$'s $y$-range or $z$-range. $\square$
However, $B$ does not contain any point in the plane $P: x - y - z = 0$.  Therefore, $B \subset \bigcup_{D\in \mathcal{A}} D$ since the union is everything but the plane.

The counter-example relies on the fact that, as defined, the boxes in $\mathcal{A}$ have open and closed brackets a certain way, and the counter-example goes a different way.  I have no doubt that if you allow more kinds of boxes in $\mathcal{A}$, while still omitting the plane $P$, then your conjecture should be true i.e. no counter-example can exist.  In fact I think the proof should be conceptually straightforward (but possibly tedious) via a continuity argument.

Added Dec 29: I've been thinking about the new $\mathcal{A}_1, \mathcal{A}_2$ in your comments.  I think the continuity argument would go like something like this: Define function $f(x,y,z) = x - y - z$.  Then it should be easy (but perhaps tedious due to the open/closed brackets case analysis) to show all the following:

*

*$\bigcup_{D\in \mathcal{A}} D$ is the complement of the plane $P: f(x,y,z) = 0$.


*Every box $B$ (whether it satisfies 1-2 or not) can be classified into one of the following 3 cases:

*

*If $f(p) > 0$ for every point $p = (x,y,z) \in B$, then $B$ is a subbox of $A \in \mathcal{A}_1$ with $a = y_2, b = z_2$ and appropriate open/closed brackets.


*Similarly, if $f(p) < 0$ for every point $p = (x,y,z) \in B$, then $B$ is a subbox of some $A \in \mathcal{A}_2$ with $a = y_1, b = z_1$ and appropriate open/closed brackets.


*Finally, if there are points $p, q \in B$ s.t. $f(p) \ge 0, f(q) \le 0$, then by continuity of $f$ there also exists point $r \in B$ s.t. $f(r) = 0$, e.g. by considering $f$ values along the line segment from $p$ to $q$.  But then $r \in P$ and therefore $B$ cannot be covered by $\bigcup_{D\in \mathcal{A}} D$.
