If $|Q|=n$, the number of subsets with at most $2$ elements is the sum of the number of subsets with $0, 1$ and $2$ elements:
$$\bigl|\{x ∈ \mathscr P(Q): |X| \le 2\}\bigr| = 1+\binom n1+\binom n2=1+\frac{n(n+1)}2,$$
and I don't see why $\dfrac{n(n+1)}2$ cannot be even. This depends on $n\equiv 0, 3\mod 4$ or not.
If $|Q|=n$, the number of subsets with at most $2$ elements is the sum of the number of subsets with $0, 1$ and $2$ elements: $$\bigl|\{x ∈ \mathscr P(Q): |X| \le 2\}\bigr| = 1+\binom n1+\binom n2=1+\frac{n(n+1)}2,$$ and I don't see why $\dfrac{n(n+1)}2$ cannot be even. This depends on $n\equiv 0, 3\mod 4$ or not.