Proving that the given set has at most $2^{n-1}$ elements
I'll expand on one of bof's comments. If $n>0$ fix some $a\in X$ and pair the subsets of $X$ as $y,\,y\cup\{a\}$ with $a\notin y$. Not both of these are in $F$ because $y\Delta(y\cup\{a\})=\{a\}$, so $|F|$ is at most half the number of subsets of $X$, i.e. $2^{n-1}$ as desired. (I'll leave you to consider the case $n=0$ separately.)