Let $F = \{E_{1}, E_{2}, \ldots, E_{s}\}$ be a family of subsets with $r$ elements of some set $X$.
Show that if the intersection of any $r+1$ (not necessarily distinct) sets in $F$ is nonempty, then the intersection of all sets in $F$ is nonempty.
Note that this has a solution but this statement seems untrue to me. But here is the given proof.
Given Proof
We assume the contrary that the intserction of all sets in $F$ is empty. Now consider the set $E_{1} = \{x_{1}, x_{1}, \ldots, x_{r}\}$. Because none of the $x_{i}, i = 1,2, \ldots, r,$ lies in the intersection of all the $E_{j}$ (this intersection being empty), it follows that for each $i$ we can find some $E_{{j}_{i}}$ such that $x_{i} \notin E_{{j}_{i}}$. Then
\begin{align}
E_{1} \cap E_{{i}_{1}} \cap E_{{i}_{2}} \cap \ldots \cap E_{{i}_{r}} = \emptyset
\end{align}
Since, at the sametime, this intersection is included in $E_{1}$ and does not contain any element of $E_{1}$. But this contradicts the hypothesis. It follows that our initial assumption was false, and hence the sets from the family $F$ have a nonempty intersection.
My Argument
This becomes a problem given that the intersection of any $r+1$ sets in $F$ are not necessarily distinct. Thus if we take $E_{1}$ and intersect with itself $r+1$ times, the intersection is not empty. However, we note that it would be possible to come up with a set $F$ where $E_{1}, \ldots, E_{s}$ do not intersect for a given $X$. And thus provides a counter example to the problem.
Comments
I don't think their proof is correct either given that we are picking $E_{{i}_{r}}$ to make sure the set intersection is empty. This is an invalid choice because it does not fit the narrative of the original question of being not necessarily distinct intersection.
Any insight or help into understanding this proof would be helpful. Note this problem comes from Putnam and Beyond as an example in the first section.
Best Answer
The proof is fine. It shows that if $\bigcap F=\varnothing$, there must necessarily be at least one family of $r+1$ or fewer members of $F$ whose intersection is empty, contradicting the hypothesis on $F$. The fact that there will still be some families of $r+1$ or fewer members of $F$ whose intersections are non-empty (e.g., all one-member families) is irrelevant: the fact that there is at least one ‘bad’ family of at most $r+1$ members of $f$ — i.e., at least one with empty intersection — is enough to contradict the hypothesis and thereby show that $\bigcap F$ cannot be empty after all.