Sum over even $k$

Sum. $\sum_k\binom{n}{k}=2^n$ and $\sum_k(-1)^k\binom{n}{k}=0$.

Compute even. Add: $2\sum_{k\text{ even}}\binom{n}{k}=2^n$, so $\sum_{k\text{ even}}\binom{n}{k}=2^{n-1}$.

Conclusion. Half the subsets have even size.