First let's define what an n-interval is: $$I = \left[a_1, ..., b_1\right] \times ... \times \left[a_n, ..., b_n\right] \subseteq \mathbb{R}^n$$ Now suppose we have $I_1, \dots, I_m$ n-intervals with pairwise disjoint interiors s.t. $I_1 \cup \dots \cup I_m = I$ n-interval.
My conjecture is that we can find a pair of index $1 \leq i < j \leq m$ such that $I_i \cap I_j$ only differ by a single "coordinate", that is there $\exists! k$ s.t. $a^i_k < b^i_k \leq a^j_k < b^j_k$.
I could prove for the case $m = 2$: Suppose there doesn't exists such $k$ then the two n-intervals are the same, impossible because they have disjoint interiors by hypothesis. Suppose now that $k$ is not unique, that is there $\exists z \neq k$ s.t. $a^1_z < b^1_z \leq a^2_z < b^2_z$: then there exists a point with $x$ s.t. $x_z = a^1_z, x_k = b^2_k$ which is not in $I_1 \cup I_2 = I$: but if this is true then $I$ is not an n-interval because it contains a point $y^1$ s.t. $y^1_z = b^2_z$ and a point $y^2$ s.t. $y^2_k = a^1_k$ but not $x$.
My problem is with the $m > 2$ case.
I tought that if one can prove that if $I = I_1 \cup \dots \cup I_n = I_1 \cup I^\star$ with $I^\star = I_2 \cup \dots \cup I_n$ n-interval then it simply follows by induction on $m$: because $I^\star = I_2 \cup \dots I_m$ is n-interval union of $m-1$ disjoint n-intervals, by induction it mus contain two n-intervals $I_i \neq I_j$ which have the property written above.