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 $1/n < a$.
You can go much easier on this. Pick any three numbers $a\_1,a\_2$ and $a\_3$ in $(0,1)$ and let
$$F = \{\{a\_1, a\_2\}, \{a\_1, a\_3\}, \{a\_2, a\_3\}\}$$
The family is finite and indexed by $\{1,2,3\}$, $F\_1\cap F\_2 = \{a\_1\}, F\_1\cap F\_3 = \{a\_2\}, F\_2\cap F\_3 = \{a\_3\}$ but
$$\bigcap\_{i=1}^3 F\_i = \emptyset$$

This example is minimal in the sense that if the index set is any smaller ($2$ sets), it doesn't work anymore and if the sets are any smaller ($1$ element per set), it doesn't work either.

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For a general construction prescribed a finite index set $I \subset \mathbb N$ with $|I| \ge 3$ you can simply pick an indexed set $A$ and define
$$A(S) := \{a\_s \mid s\in S\}\\
F\_i := A(I \setminus \{i\})$$
Then $F\_i\cap F\_j = A(I\setminus\{i,j\}) \ne \emptyset$ since $|I| > 2$ and
$$\bigcap\_{i\in I} F\_i = A\left(I \setminus \bigcup\_{i\in I} \{i\}\right) = \emptyset$$