Compressing (n+1)-bit input to n-bit output makes the average output correspond to 2^(n+1)/2^n = 2 inputs. By Dijkstra's rephrasing of the pigeonhole principle, the maximum number of inputs colliding on the same output must be at least 2.
Or we could make it more robust and say we're compressing to n-bit-or-smaller output. Then the average output corresponds to 2^(n+1)/(2^(n+1)-1) inputs. That average is still strictly greater than 1, so the maximum number of inputs colliding on the same output must also be strictly greater than 1 (at least 2, since that's the smallest natural which is greater than 1).
Say you're trying to find a way to compress every n+1 bit string into an n bit string. Losslessness means that you have to come up with a function d from n bit string to n+1 bit string that decompresses. The number of n+1 bit string is larger, so there must be n+1 bit strings that are not in the image of d, which means that such a function d cannot exist.
We might as well call this "the whole-pigeon principle."
When you compress a file, you are representing a big number with a small number.
There are more big numbers than there are small numbers.
Specifically, they were looking for how you write the can't-always-compress proof using formulation (1)