Binomial coefficient bound

Sizes. There are $n+1$ possible sizes.

No pigeonhole needed if we choose $n+2$ subsets (two same size by pigeonhole).

But with $\binom{n}{\lfloor n/2\rfloor}+1$ subsets: if all sizes are distinct, at most $n+1$ sizes. So we'd need $n+1\ge\binom{n}{\lfloor n/2\rfloor}+1$, which is false for $n\ge4$. Contradiction.

Conclusion. Among $\binom{n}{\lfloor n/2\rfloor}+1$ subsets, two have the same size (for $n\ge4$).