Suppose we have a family $F$ such that:

- For each $A \in F$ we have $|A| = k$ and $A \subset n$.
- For each $A,B \in F$ we have $A \cap B \neq \emptyset$.

It is easy to show that there exists a nonempty set $B \subset n$ such that:

- For each $A \in F$ we have $B \cap A \neq \emptyset$.
- $|B| \le k^2$.

*Proof*: Say $|F| = m$. Define $d(x) = |\{A \in F| x \in A\}|$. Define $D(A) = \sum_{x \in A}d(x)$. In order to have condition (2) for all $A \in F$ we must have $D(A) \ge m$. Thus $A$ must contain at least one element $x$ such that $d(x) \ge m/k$. But there are at most $k^2$ such elements, because $D(\bigcup F) = \sum_{x < n} d(x) = km$. Hence we can take $B=\{a \in n| d(a) \ge m/k \}$.

** Question**: The above result gives an upper limit on the size of $B$ that is not dependent on $n$. Is this still true for $A \subset \omega$ and $F$ infinite?

The same question was asked in https://math.stackexchange.com/questions/1948282/unavoidable-finite-set-for-infinite-intersecting-family, but got no reply.