Can $[0,1]^4$ be partitioned into copies of $(0,1)^3$? Is there a partition of $[0,1]^4$ such that every member of the partition is homeomorphic to $(0,1)^3$?
More generally, I would like to know for which values of $m$ and $n$ there is a partition of $[0,1]^n$ into topological copies of $(0,1)^m$.
For such a partition to exist, it is necessary that $m < n$.
Another simple observation is that if $[0,1]^n$ can be partitioned into copies of $(0,1)^m$, then $[0,1]^{n+k}$ can be partitioned into copies of $(0,1)^m$ for all $k$. (This is because $[0,1]^{n+k}$ can be partitioned into copies of $[0,1]^n$, each of which can then be further partitioned into copies of $(0,1)^m$.)
I have solved the two simplest instances of this general problem already:
$\bullet$ $[0,1]^2$ can be partitioned into copies of $(0,1)$ (shown below), and this implies $[0,1]^n$ can be partitioned into copies of $(0,1)$ for all $n > 1$, and
$\bullet$ $[0,1]^3$ can be partitioned into copies of $(0,1)^2$, and this implies $[0,1]^n$ can be partitioned into copies of $(0,1)^2$ for all $n > 2$.
The title question is the simplest instance of this more general problem for which I don't already know the answer.
A partition of $[0,1]^2$ into copies of $(0,1)$:

The first three pictures in the bottom row show three partition pieces. The last picture is what's left over after removing these three, and because the leftover bit is simply four copies of $(0,1)^2$ (up to homeomorphism), it's easy to partition the leftovers into copies of $(0,1)$.
To get a partition of $[0,1]^3$ into copies of $(0,1)^2$, take the partition of $[0,1]^2$ into copies of $(0,1)$ illustrated above, and rotate everything about the vertical line through the center of the square, in the same plane as the square (the axis of symmetry for the M-shaped bit of the partition). Upon rotation, our square transforms into a cylinder, which is (homeomorphic to) $[0,1]^3$. Each of the first three partition pieces transforms into a copy of $(0,1)^2$. The leftovers in the rightmost picture transform into two copies of $(0,1)^3$ plus one copy of $S^1 \times (0,1)^2$, and it is easy to further partition these into copies of $(0,1)^2$.
The same trick does not seem to work for stepping up another dimension to $(0,1)^3$ and $[0,1]^4$.
Lastly, let me mention that if we reverse the roles of $(0,1)$ and $[0,1]$ in my general question, then the answer is known (and not too difficult to prove):
there is a partition of $(0,1)^n$ into topological copies of $[0,1]^m$ if and only if $m < n$. (This is proved as Theorem 2.2(i) in this 1979 paper of Bankston and McGovern.)
 A: EDIT: As Will pointed out, the attempt here doesn't quite work. I've corrected the false claims and added a caveat for the missing ingredient.
If I'm not mistaken, $[0,1]^{n+1}$ can be partitioned into homeomorphic copies of $(0,1)^n$ via the following procedure:

*

*Regard $[0,1]^{n+1}$ as $$\{(x_1,\ldots,x_{n+1}):x_{n+1}\geq 0\}\cup\{\infty\}$$

*The first $(0,1)^n$ slice is $$\{(x_1,\ldots,x_n,0):x_1^2+\ldots+x_n^2> 1\}\cup\{\infty\}$$

*The second slice is $$\{(x_1,\ldots,x_n,0):x_1^2+\ldots+x_n^2\leq 1\}$$ union the "upper half-torus" parallel to the $x_{n+1}=0$ plane of inner radius 1 and outer radius 2

*After removing these two slices, what's left is homeomorphic to $S^{n-1}\times (0,1)^2$ and $(0,1)^{n+1}$ (which can be sliced via the pieces $\{t\}\times(0,1)^n$).

*By thickening in the extra two dimensions, it suffices to show that $S^{n-1}$ can be sliced by copies of $(0,1)^{n-2}$.

Here it remains to be demonstrated that $S^k$ can be sliced by $(0,1)^{k-1}$; inductively, one can assume that $[0,1]^d$ can be sliced by $(0,1)^{d-1}$ for all $d=1,\ldots,k+1$.
A: Almost exactly the same picture can be drawn for the 2-dimensional and 3-dimensional cases (simpler than the OP's) and then it's easy to generalize to higher dimensions by induction. Here $I$ indicates the open interval and $\bar{I}$ the closed one.

The general case $n\ge 4$ is now easy: as above one can decompose ${\bar I}^n$ as follows: $I^{n-1}\sqcup I^{n-1}\sqcup I^n \sqcup S^{n-2}\times I^2$ and $S^{n-2}$ itself can be decomposed as union of copies of $I^{n-3}$ by inductively using the decomposition of $\bar{I}^{n-2}$ joined to its complement in $S^{n-2}$ which is another copy of $I^{n-2}$ and therefore also a union of copies of $I^{n-3}$.
NOTE: I didn't fully understand Ben Johnsrude's earlier answer, but I think that mine may be essentially the same as his.
