How to prove that $|\mathcal{F}|\leq 2^{n-1}$?
HINT: Write as $V$ the $n$-element set. Then there are $2^n$ subsets of $V$, and furthermore those $2^n$ subsets of $V$ can be partitioned into $2^{n-1}$ pairs $\{S,\bar{S}\}$, where $\bar{S}$ is the complement of $S$. If $\cal{F}$ is intersecting however, then for each $S \in \cal{F}$, its complement $\bar{S}$ in $V$ is not in $\cal{F}$, equivalently, only one of $S$, $\bar{S}$ can be in $\cal{F}$.