Exactly half of the elements of $\mathcal{P}(A)$ are odd-sized

Fix an element $a\in A$ (this is the point where $A\ne\emptyset$ is needed). Then $$S\mapsto S\operatorname{\Delta}\{a\}$$ (symmetric difference) is a bijection from the set of odd subsets to the set of even subsets.

Hint: One can prove this by induction on the size of $A$. Assume it was true for sets of size $n$ and let $A=\{a_1,\ldots,a_{n+1}\}$. Then every subset of $A$ is either a subset of $\{a_1,\ldots,a_n\}$ or it is a copy of such subset with the addition of $\{a_{n+1}\}$. Use the induction hypothesis to conclude that the sets which do not contain $a_{n+1}$ have this property (with respect to $\{a_1,\ldots,a_n\}$, by adding $a_{n+1}$ you send exactly the same number of odd sets to even size sets, and vice versa; therefore the ratio remains true for $A$.

For $n\in\Bbb Z^+$ let $[n]=\{1,2,\dots,n\}$. Clearly $[1]$ has one even subset and one odd subset. Suppose that $[n]$ has equal numbers of odd and even subsets for some $n\in\Bbb Z^+$. The even subsets of $[n+1]$ are of two types:

• the even subsets of $[n]$; and
• the sets of the form $A\cup\{n+1\}$, where $A$ is an odd subset of $[n]$.

By the induction hypotheses there are the same number of sets of the second type as there are of the first, so $[n+1]$ has twice as many even subsets as $[n]$. But $[n+1]$ also has twice as many subsets altogether as $[n]$, so it must have twice as many odd subsets as well, which clearly implies that it has equal numbers of odd and even subsets.