Proving that 6-1 has only one solution is a pretty difficult challenge. I've tried to put down into words how I proved this. Hopefully, it makes sense! I broke up my thinking into six steps. If you want to prove it yourself, only read down as far as you need to.
Explanation below (spoilers):
1. Even without considering limitations on movement, the "lying planet" can be narrowed down to either of the middle two planets in the second row (the green or brown).
2. Considering each planet as the liar individually, there are two solutions with a brown liar and only one with a green liar, for a total of three possible solutions. Now it needs to be proved that only one of these is possible. (In the following diagrams, an H is meant to be two vertical lines. The 3 is the green planet and the 6 is the brown planet).
~~Possibility 1~~
| H
=3 6
H
~~Possibility 2~~
| H
-3-6
|
~~Possibility 3~~
| H
-3=6
H
3. Horizontal and vertical movements are effectively separate in that a horizontal move will never change vertical lines, and vice versa. Only the vertical movement needs to be considered here.
4. The key is proving that the aforementioned brown planet (Row 2, Column 3) is limited to only connecting to the planet below it with one line, not two. If this is proved, only the second of the above possibilities would be valid.
5. It can be proved (by considering parity) that within the movement constraints and possible relative locations of the two astronauts in level 6-1, a single line, by itself, can never be created.
6. Given that (again, only considering the vertical lines) the puzzle CAN be solved using the second of the previous three possibilities, the other two must not be possible because the difference in vertical lines is exactly one line (below the lying brown planet).