This comes from baby Rudin's definition 11.4, and it is left to the reader to verify the above. Elementary sets are defined as the finite union of any open/closed/half-open k-cells.
As it was left to the reader, I thought this would require a rather short proof, so I'm not sure if my attempt was the simplest way to do it. Also, I'm not sure of the correctness of my proof.
Let $F$ denote the family of elementary sets on $\mathbb{R^n}$ and let $A$, $B \in F$ s.t. $A = \cup_{i=1}^{n}A_i$ and $B = \cup_{j=1}^{m}B_j$ where $A_i, B_j$ are k-cells.
Clearly, $A\cup B \in F$, so it remains to prove that $A-B\in F$. $A-B = \cap_{1\leq i\leq n, 1\leq j\leq m}A_i-B_j$, hence it is sufficient to show $A_i-B_j\in F$.
Let the 2 "opposite corners" of $A_i$ be $a_1, a_2$ and $B_j$ be $b_1, b_2$. We assume that $B_j$ intersects $A_i$ at it's corner $a_2$, such that only $b_1$ lies in $A_i$. If the intersection is at any other "corner" of $A_i$, we use a symmetry argument to generalise. If $A_i \cap B_j$ lies within $A_i$, such that $a_1 \leq b_1$ and $b_2 \leq a_2$, then we apply the subsequent argument 2 times, once removing the k-cell with corners $a_1,b_2$ from $A_i$, then removing $B_j$ from the k-cell with corners $a_1,b_2$. Now let the components of $b_1$ be $(c_1,c_2,...,c_n)$ and components of $a_2$ be $(d_1,d_2,...,d_n)$. Going back to the basic case where the intersection lies on the corner of $A_i$, consider the finite union of k-cells with corners $a_1,(f_1,f_2,...,f_n)$, where $f_i$ is either $c_i$ or $d_i$ such that not all $f_i = d_i$.
This union is equal to $A_i-B_j$.