Set of sets such that any three of them has nonempty intersection. Show all of them have nonempty intersection. 
Given $2^{n-1}$ subsets of a set with $n$ elements with the property that any three have nonempty intersection, prove that the intersection of all the sets is nonempty. 

My attempt:
Let $|X|=n$ and $S \subset \mathcal{P}(X)$ be the above mentioned set. If the intersection of all the sets in $S$ is nonempty and contains W.L.O.G $a \in X$, then the set of all subsets of $X$ containing $a$ satisfy the requirement. 

 A: The claim clearly holds for $n=1$ and $n=2$.  (See Hagen von Eitzen's comment below.  With a different interpretation of the problem statement, the claim is false for $n=1$ and $n=2$.)  In what follows, we assume that $n\geq 3$.  
Let $S$ be a set with $n$ elements and $\mathcal{F}$ a family of subsets of $S$ with the required properties.  That is, $\mathcal{F}$ satisfies the condition that $|\mathcal{F}|=2^{n-1}$ and that any three (not necessarily distinct) sets in $\mathcal{F}$ have a nonempty intersection.  
Consider the vector space $V:=\mathbb{F}_2^S$ with basis $\big\{e_s\,|\,s\in S\big\}$ over the field $\mathbb{F}_2$ with $2$ elements.  We identify each subset $X\subseteq S$ with the vector $$e_X:=\sum_{s\in X}\,e_s\,.$$
In particular, $e_{\{s\}}=e_s$ for all $s\in S$, and $e_\emptyset=0$.  Equip $V$ with the nondegenerate inner product $\langle\_,\_\rangle$ defined by
$$\left\langle e_X,e_Y\right\rangle :=|X\cap Y|$$
for every $X,Y\subseteq S$ (here, $|X\cap Y|$ is considered modulo $2$).
We claim that $W:=\mathbb{F}_2^S\setminus \big\{e_X\,\big|\,X\in\mathcal{F}\big\}$ is a vector subspace of $V$.  That is, $W$ is closed under addition and scalar multiplication.  Clearly, $W$ is closed under scalar multiplication since $0=e_\emptyset\in W$ and $\mathbb{F}_2=\{0,1\}$.  We are left with closure under addition.
For a subset $X$ of $S$, write $X^\complement$ for $S\setminus X$.  It can be easily seen that, for each $X\subseteq S$, either $X\in\mathcal{F}$ or $X^\complement \in \mathcal{F}$, but not both.  
Suppose $A$ and $B$ are subsets of $S$ such that $e_A\in W$ and $e_B\in W$.  Then, $A^\complement\in\mathcal{F}$ and $B^\complement \in\mathcal{F}$.  Now, $e_A+e_B=e_{A\triangle B}$, where $\triangle$ is the symmetric difference.  Because $$(A\triangle B)\cap A^\complement \cap B^\complement=\emptyset\,,$$ $A\triangle B$ cannot be in $\mathcal{F}$.  Thus, $e_{A\triangle B}\in W$.  
Consequently, $W$ is indeed a subspace of $V$ with dimension $n-1$, i.e., with codimension $1$.  There exists a unique linear functional $\varphi:V\to \mathbb{F}_2$ with kernel $W$.  Hence, $\varphi=\langle e_T,\_\rangle$ for some nonempty $T\subseteq S$ (this is because $\langle\_,\_\rangle$ is a nondegenerate symmetric bilinear form on $V$).  It remains to show that $T$ is a singleton.  
If $T$ contains at least three pairwise distinct elements, say $a$, $b$, and $c$, then take $A:=\{a\}$, $B:=\{b\}$, and $C:=\{a,b,c\}$.  Since $e_A$, $e_B$, and $e_C$ are not in $\ker(\varphi)$, $A$, $B$, and $C$ must be in $\mathcal{F}$, but $A\cap B\cap C=\emptyset$.  This is a contradiction, so $|T|\leq 2$.  
If $|T|=2$, say, $T=\{a,b\}$, then we take $c\in S\setminus T$ (noting that $|S|=n\geq 3>|T|$), and set $A:=\{a\}$, $B:=\{b\}$, and $C:=\{a,c\}$ and apply the same argument as before to arrive at a contradiction.  Ergo, $T$ is a singleton: $T=\{t\}$ for some $t\in S$.  This proves that every set in $\mathcal{F}$ contains $t$.
