Consider that every open set in $\mathbb{R}^d, d \geq 1$, can be written as a countable union of almost disjoint cubes.
One way to prove that theorem is to first construct a grid generated by the lattice of integers $\mathbb{Z}^d$, then rejecting all such cubes not completely contained in our open set, and then using grids with side lengths $2^{-n}$ obtained by successively bisecting the original grid, subsequently accepting or rejecting these new cubes depending on whether they are entirely contained in the open set, etc.
The proof itself is straight forward to me, however, how can we be sure that the lattice of integers is "shaped" the way it seems like it should be for, say, $ d = 13$? Is it enough to say that if the open set $O = (0,1) \times (0,1) \times \cdots \times (0,1)$ then $O \subset R = [0,1] \times [0,1] \times \cdots \times [0,1]$? I guess I'm asking if we define one set to contain another $\iff$ all of it's coordinate intervals contain all of the others.
We define one set to contain another if and only if every element of the latter is an element of the former. Here's how we know $O \subset R$: let $x \in O$. By the definition of $O$, $x$ is a length-$d$ sequence $(x_1,x_2,\ldots,x_d)$ where every $x_i$ is strictly between $0$ and $1$. By the definition of $R$, $R$ includes every length-$d$ sequence $(y_1,y_2,\ldots,y_d)$ where every $y_i$ is between $0$ and $1$ (possibly including $0$ and $1$). Since $x$ fits that description, $x \in R$. So every element of $O$ is in $R$, and therefore $O \subset R$.
The proof that you described is "glossing over" a lot of details, because an argument like the one I just gave is a pain to say every time. If you were to work it out in complete detail, you'd find that the argument is talking about the individual components of the points involved, boiling everything down to one point at a time - in other words, the argument would look something like "Let $O$ be an open subset of $\mathbb{R}^d$. We perform such-and-such a process. Let $x$ be an arbitrary element of $O$. $\ldots$ (etc., etc., etc.) $\ldots$ Therefore $x$ is included in one of the boxes we added on the $n$th step. So every element of $O$ is eventually covered by one of our boxes, so the union of our boxes contains all of $O$."
Importantly, this argument has nothing at all to do with the "shape" of the integer lattice - all it needs is that the integer lattice is distributed through the whole space, and that each "box" in the lattice can be subdivided into smaller "boxes".