Suppose we have a family $F$ such that:

- For each $A \in F$ we have $|A| = k$
- For each $A,B \in F$ we have $A \cap B \neq \emptyset$
- If we have $C$ such that for each $A \in F$ we get $C \cap A \neq \emptyset$ then $|C| \ge k$

** Question**: For a fixed $k$, can $F$ be arbitrarily large? Can $F$ be infinite?

*Simple observations*:
For $k = 1$ it's obvious that $|F| = 1$, namely a singleton.

For $k = 2$ we have $|F| \leq 3$, because w.l.o.g $\{1,2\} \in F$, so w.l.o.g $\{1,3\} \in F$. Working towards contradiction, assume $4 \in \bigcup F$. Then a set $\{x,4\}$ exists, but then $x=1$, otherwise $\{x,4\} \cap \{1,2\} = \emptyset$ or $\{x,4\} \cap \{1,3\} = \emptyset$. So $\{1,4\} \in F$, but then if $1 \notin A$, $A \cap \{1,2\} = \emptyset$ or $A \cap \{1,3\} = \emptyset$ or $A \cap \{1,4\} = \emptyset$, therefore $A \notin F$. Hence $\{1\}$ intersects all sets in $F$, contrary to condition (3). Therefore $|\bigcup F| \le 3$, and $\{\{1,2\},\{1,3\},\{2,3\}\}$ is the biggest $F$ possible.

For $k = 3$ the Fano plane provides a family with at least 7 sets.

Obviously it's the third condition that makes a finite limit even possible, otherwise Erdos-Ko-Rado gives us arbitrarily large intersecting families.

I uploaded the same question (although I phrased it better here) to stack overflow but got no response. https://math.stackexchange.com/questions/1942413/finite-limit-to-size-of-intersecting-family-with-no-smaller-intersecting-set