Question: Give a family of sets $\{F\_a\}$ with each $F\_a\subseteq (0,1)$ and $F\_a \cap F\_b \neq \emptyset$ but $\bigcap\_aF\_a = \emptyset$.
What is an example of such a family of sets?

Answer: 
A standard example in analysis is to let your $F\_n = (0,1/n), n \in \mathbb{N}$. This ensures that any two $F\_n$ intersect, but the intersection of all of the $F\_n$ is $\emptyset$. 

Specifically, for $n < m$, $F\_n \cap F\_m = F\_m$. However, for any $a > 0$, $F\_n$ does not contain $a$ when $1/n < a$.
Let me try.

Take the numbers $K\_2 \subset \mathbb{N}$ divisible by $2$, and in general $K\_n$ divisible by $n > 2$. 

For any $n,m$ the intersection of $K\_n, K\_m$ is non-empty containing the multiples of $n\times m$.

But the intersection $\cap\_a K\_a = \emptyset$

Now take the set $S\_2$ of the form $\frac{1}{k\_2}, k\_2 \in K\_2$, and in general $S\_n$ of the form $\frac{1}{k\_n}, n > 2$, that is, the inverse of the previous sets. 

They satisfy $\cap\_a S\_a = \emptyset$ and $S\_a \subseteq (0,1)$.