I think that what you are talking about has a cubical version for "partial boxes", and it might be interesting to relate the two versions. The following is Lemma 3.3 of

R. Brown and P.J. Higgins, ``Colimit theorems for relative homotopy
groups'', *J. Pure Appl. Algebra* 22 (1981) 11-41.

and is also given an exposition in our EMS Tract vol 15 "Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids" (2011) Section 11.3.i *Fibrant cubical sets*.

[Chains of partial boxes] Let $B$, $B'$ be partial
boxes in an $r$-cell $C$ of $\mathbb I^n$ such that $B' \subseteq
B.$ Then there is a chain
$$B = B_s \searrow B_{s-1} \searrow \cdots \searrow B_1 = B'$$
such that

(i) each $B_i$ is a partial box in $C$;

(ii) $B_{i+1} = B_i \cup a_i$ where $a_i$ is an $(r - 1)$-cell of $C$ not in $B_i$;

(iii) $a_i \cap B_i$ is a partial box in $a_i.$

The definition of a partial box is as follows:

Let $C$ be an $r$-cell in the $n$-cube $\mathbb I^n.$ Two $(r -
1)$-faces of $C$ are called *opposite*
if they do not meet (except possibly in degenerate elements). A *partial $(r-1)$--box* in
$C$ is a subcomplex $B$ of $C$ generated by one $(r - 1)$-face $b$
of $C$ (called a *base* of $B$) and a number, possibly zero, of
other $(r - 1)$-faces of $C$ none of which is opposite to $b$. The
partial box is a *box* if its $(r - 1)$-cells consist of all but
one of the $(r - 1)$-faces of $C$.

I find the cubical arguments more geometric and so easier for me to find and to follow. The above Lemma is a crucial part of the whole theory. It may be related to some results in Kan's first paper.

There are other reasons for working with cubical sets, mainly the idea of "algebraic inverses to subdivision", using cubical sets with compositions, and also the rule $I^m \times I^n =I^{m+n}$.