Prove Intersection of Large Enough Subsets is Non-empty I am trying to prove the following claim. I did not find the claim in a book, but I believe it to be true.

Claim: Consider a finite set $P\neq \emptyset$ and subsets $S_1, S_2, S_3 \subset P$, each of size
  $s$. Suppose $s > \frac{2}{3} |P|$. Prove that $S_1 \cap S_2 \cap S_3 \neq \emptyset$.

Here is my proof.

Proof: Consider the elements of $S_1$ and $S_2$. There must be at minimum $\left\lfloor \frac{2|P|}{3} \right\rfloor + 1$ elements in
  each set. Therefore  $$\begin{align*} |P - S_1| &= |P| - s \\ &\le |P|
 - \left\lfloor \frac{2|P|}{3} \right\rfloor - 1 \\ &= \left\lceil \frac{|P|}{3} \right\rceil - 1  \end{align*}$$ Thus $S_1$ and $S_2$ can have at most $\left\lceil \frac{|P|}{3} \right\rceil - 1$ distinct
   elements. The other elements must belong to their intersection.
   $$\begin{align*} |S_1 \cap S_2|  &\ge \left\lfloor \frac{2|P|}{3}
 \right\rfloor + 1 - \left(\left\lceil \frac{|P|}{3} \right\rceil -
 1\right)\\ &\ge \left\lfloor \frac{2|P|}{3} \right\rfloor - 
 \left\lfloor \frac{|P|}{3} \right\rfloor + 1\\ &=  \left\lfloor
 \frac{|P| + 1}{3} \right\rfloor + 1 \end{align*}$$
Thus, 
  $$\begin{align*} |S_1 \cap S_2 \cap S_3| &= |S_1 \cap S_2| +
 |S_3| - |(S_1 \cap S_2) \cup S_3| \\ &\ge \left\lfloor \frac{|P| + 1}{3} \right\rfloor + 1 + \left\lfloor \frac{2|P|}{3} \right\rfloor + 1 - |P|\\ 
&>0 \end{align*}$$

I am looking for verification, criticism, and for alternative proofs.
 A: Here is how I would prove it. 
Count the total membership count, that is, the total number of times it happens that $p\in S_i$, for any point $p\in P$ and any of the sets $S_i$. Since each set $S_i$ is more than 2/3 of the points in $P$, this total membership count is strictly more than $3\cdot \frac 23|P|$. Thus, the total membership count is strictly more than $2|P|$. 
But if no point $p$ is in all three sets, then every point is in at most two $S_i$, and so the total membership count would be at most $2|P|$, a contradiction. 
This way of arguing shows the stronger result: you only need one of the sets to have strictly more than 2/3 of the points, if the other two sets have at least 2/3 of the points.
The argument generalizes to $k$ sets, provided that each of them has fraction $\frac{k-1}k$ of the total, with at least one having strictly more than this. In the case of two sets, this is the simple assertion that if $A$ has at least half the points and $B$ has more than half, then $A\cap B$ is not empty.
