There is a unique solution, and it can be found by logic alone (i.e., human-level logic that doesn't require backtracking of more than a few steps). A few things to note that help with the process of elimination:
- Any two boxes connected to each other by a solid line contain subsets differing in size by exactly one; and so any two boxes connected to each other by a solid path with an even (odd) length contain subsets differing in size by an even (odd) number.
- All pairs of boxes joined by a dotted line are separated by an even-length solid path, and hence differ in size by an even number of elements, except the pair in the lower left.
- The subset $\{1\}$ can't lie on a dotted line, since it's the unique subset with the highest value $1$.
- The subsets $\{1,2\}$ and $\{2\}$ have the property that if either of them is on a dotted line, it's connected to the other by that line, since they're the unique subsets with the highest value $2$. By fact 2., since they differ in size by an odd number of elements, they either both go in the pair of boxes on the lower left, or neither is on a dotted line.
By fact 1., the subset $\{1\}$ needs to go in one of the four boxes not on any dotted line. Three of those boxes are part of solid-solid-dotted triangles: the subset $\{1\}$ can't go at the solid-solid vertex of any of those, either, because its neighbors would have to be of the form $\{1,a\}$ and $\{1,b\}$, with $a=b$ (because of the dotted line) and $a\neq b$ (because they are different boxes): a contradiction. That leaves the upper right-hand corner as the unique location for $\{1\}$.
By fact 4., $\{2\}$ and $\{1,2\}$ either (a) both go in the pair of boxes on the lower left, or (b) neither touches a dotted line. If it's (b), then there are three places left that they can go: the two boxes on the lower right, and the box next to $\{1\}$ in the upper right. They can't both go in the two boxes on the lower right (even-solid-distance), so one must go next to $\{1\}$... that would have to be $\{1,2\}$ because of the solid connection to $\{1\}$. But $\{1,2\}$'s other solid-line neighbors would have to be $\{1,3\}$ and $\{1,4\}$, which don't have the same highest element, and so (b) is a contradiction. Therefore one of $\{2\}$ and $\{1,2\}$ goes on the extreme lower left: that's an even-solid-distance from the upper right, which contains $\{1\}$, so $\{2\}$ must be on the lower left. And $\{1,2\}$ is just above it.
All the remaining boxes and dotted lines are in solid-solid-dotted triangles (some are in more than one). If $\{1,3\}$ or $\{2,3\}$ are on a dotted line, they must be connected to the other by it (only two even-size subsets with the highest value $3$), and the remaining vertex in the solid-solid-dotted triangle must be $\{1,2,3\}$ or $\{3\}$. They also can't go in the boxes on the lower right (even-solid-distance from the upper right), leaving only one box not on a dotted line, so they must be on a dotted line. It can't be the long dotted line (no other highest-number-$3$ and even-size partner), or either of the two short dotted lines an even-solid-distance from the upper right. That leaves the upper left corner and the box to its southeast for $\{1,3\}$ and $\{2,3\}$, and forces $\{1,2,3\}$ into the box above $\{1,2\}$ (since the other choice, $\{3\}$, can't be joined by a solid line to $\{1,2\}$). Now $\{3\}$ must be to its southeast, since it's the only highest-number-$3$ partner left.
The upper left can't be $\{2,3\}$, because the only ways to get from $\{1\}$ to $\{2,3\}$ with three solid lines use $\{1,2\}$ or $\{1,3\}$, which aren't available; so it must be $\{1,3\}$, and the other boxes in the top row must then be $\{1,4,3\}$ and $\{1,4\}$ (left to right).
The rest can now be filled in rapidly. $\{2,4\}$ to the right of $\{2\}$. $\{3,4\}$ to the right of $\{3\}$. $\{1,2,3,4\}$ to the northeast of $\{3,4\}$ (only even-sized subset left). $\{2,3,4\}$ connecting $\{3,4\}$ to $\{1,2,3,4\}$ by solid lines. $\{4\}$ connecting $\{2,4\}$ and $\{3,4\}$ by solid lines. And $\{1,2,4\}$ connecting $\{1,4\}$ and $\{1,2,3,4\}$ by solid lines. And done.