Half-open boxes are closed under intersection, and the complement of a half-open box is a finite disjoint union of half-open boxes

For claim 1, a point $x\in {\mathbb{R}}^n$ lies in $B(a,b)\cap B(a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ exactly when $a_i<x_i\le b_i$ and $a_i^{\mathrm{\prime }}<x_i\le b_i^{\mathrm{\prime }}$ for every $i<n$; the order being total, each two-element set $\{a_i,a_i^{\mathrm{\prime }}\}$ has a greatest member $c_i$ and each $\{b_i,b_i^{\mathrm{\prime }}\}$ a least member $d_i$, and for a real $x_i$ the conjunction $a_i<x_i$ and $a_i^{\mathrm{\prime }}<x_i$ says exactly $c_i<x_i$ while $x_i\le b_i$ and $x_i\le b_i^{\mathrm{\prime }}$ says exactly $x_i\le d_i$, so the intersection is $B(c,d)$; iterating along a list of length $m$ gives the finite case by induction on $m$, with the empty list giving ${\mathbb{R}}^n=B(-\mathbf{\infty },+\mathbf{\infty })$.

For claim 2 in the degenerate case, if $B(a,b)=\mathrm{\varnothing }$ then ${\mathbb{R}}^n\setminus B(a,b)={\mathbb{R}}^n=(-\mathrm{\infty },+\mathrm{\infty }{]}^n$, a list with the single member ${\mathbb{R}}^n$, whose members are vacuously pairwise disjoint.

For claim 2 in the remaining case, assume $B(a,b)\mathrm{\ne }\mathrm{\varnothing }$, so $a_i<b_i$ for every $i<n$, and for $i<n$ define two parameter pairs $(a^{i,0},b^{i,0})$ and $(a^{i,1},b^{i,1})$ by setting, in coordinates $j<i$, $a_j^{i,ϵ}:=a_j$ and $b_j^{i,ϵ}:=b_j$; in coordinate $i$, $(a_i^{i,0},b_i^{i,0}):=(-\mathrm{\infty },a_i)$ and $(a_i^{i,1},b_i^{i,1}):=(b_i,+\mathrm{\infty })$; and in coordinates $j>i$, $a_j^{i,ϵ}:=-\mathrm{\infty }$ and $b_j^{i,ϵ}:=+\mathrm{\infty }$.

Still for claim 2, every $x\mathrm{\notin }B(a,b)$ lies in one of these $2n$ boxes: the set of $i<n$ with $\mathrm{\neg }(a_i<x_i\le b_i)$ is a nonempty subset of $n$, so it has a least member $i$; then $a_j<x_j\le b_j$ for every $j<i$, and by totality either $x_i\le a_i$, putting $x$ in $B(a^{i,0},b^{i,0})$, or $x_i>b_i$, putting $x$ in $B(a^{i,1},b^{i,1})$, the coordinates $j>i$ being unconstrained in both.

Still for claim 2, each of the $2n$ boxes is disjoint from $B(a,b)$, since its points satisfy $x_i\le a_i$ or $x_i>b_i$; and two of them are disjoint from one another, because for $i<i^{\mathrm{\prime }}$ a point of a box with index $i$ fails $a_i<x_i\le b_i$ while a point of a box with index $i^{\mathrm{\prime }}$ satisfies it, and for a common $i$ a point of both would satisfy $b_i<x_i\le a_i$, contradicting $a_i<b_i$.

Claim 1 is step 1.1, and claim 2 is step 1.2 in the empty case and steps 2.1 and 2.2 in the nonempty case, the union of the $2n$ boxes being exactly ${\mathbb{R}}^n\setminus B(a,b)$.