We induct on $n$. The base case $n=1$ is clear. So now $n\rightarrow n+1$. WLOG let $X=[n+1]$ and $X'=[n]$. Moreover, note that $$|\mathcal{F}| > \sum_{i=0}^{k-1}{n+1 \choose i} = 2\left[\sum_{i=0}^{k-2}{n \choose i}\right] + {n \choose k-1} \, .$$Let all sets $V \in\mathcal{F}$ for which $n+1 \not\in V, V\cup\{n+1\} \in \mathcal{F}$ be in $\mathcal{V}$. Similarly, all other sets are in $\mathcal{F}\backslash\mathcal{V}:=\mathcal{U}$. Suppose that $|\mathcal{U}|\leq \sum_{i=0}^{k-1} {n \choose i}$, otherwise the result follows by induction (by noting $U\backslash\{n+1\}\subseteq X', U\in \mathcal{U}$ and for any $A,B \in \mathcal{U}$, we have $A\backslash\{n+1\}\neq B\backslash \{n+1\}$). Thus $|\mathcal{V}|=|\mathcal{F}|-|\mathcal{U}|>\sum_{i=0}^{k-2}{n \choose i}$. By induction, $\mathcal{V}$ contains a set $Y' \subseteq X', |Y'|=k-1$ with $p(Y')=Y' \cap \mathcal{V}$. Now consider $Y = Y' \cup \{n+1\}$. For a set $S\in p(Y')$, we can find $V \in \mathcal{V}\subseteq \mathcal{F}$ s.t. $S=V \cap Y'=V\cap Y$ by above. So consider $S \in p(Y)\backslash p(Y')$. Clearly $n+1 \in S$ and for $S\backslash \{n+1\}$ we can again find $V \in \mathcal{V}$ s.t. $S\backslash \{n+1\} = V \cap Y'$. In fact, by how we defined $\mathcal{V}$, we can find $U \in \mathcal{U} \subseteq \mathcal{F}$ s.t. $S =(S\backslash\{n+1\})\cup \{n+1\}= (V\cup \{n+1\})\cap (Y' \cup \{n+1\})= U \cap Y$. This shows that $Y$ indeed satisfies $Y\subseteq X, |Y|=k$ and $p(Y)=\mathcal{F}\cap Y$, which completes the problem. $\square$