Theorem:
Suppose that $A_1, . . . , A_k$ is a collection of $k ≥ 2$ sets. Show that (using induction),
$$\big| \bigcup\limits_{i=1}^{k}A_i \big| \ge \sum\limits_{i=1}^{k} \big|A_i| - \sum\limits_{\{i,j\}} \big|A_i \cap A_j \big| $$
where the second term on the right sums over all subsets of [k] of size 2.
Question:
Deduce that in every collection of 5 subsets of size 6 drawn from $\{1, 2, . . . , 15\}$, at least two of the subsets must intersect in at least two points.
What I've tried:
Let $|I| = 6$ and $i$ be one of 5 subsets s.t.
$$\big| \bigcup\limits_{i=1}^{k}A_i \big| = \sum_{0 \neq I \subseteq [k]}(-1)^{6 + 1} \big| \bigcap\limits_{i \in I} A_i \big|$$
Since, 6 does not divide 15,
$$\big| \bigcap\limits_{i \in I} A_i \big| = 0$$
$$\implies \big| \bigcup\limits_{i=1}^{k}A_i \big| = 0$$
Since, RHS of the inequality cannot be negative as
$$\sum\limits_{i=1}^{k} \big|A_i| > \sum\limits_{\{i,j\}} \big|A_i \cap A_j \big|$$
We are left with,
$$\sum\limits_{i=1}^{k} \big|A_i| - \sum\limits_{\{i,j\}} \big|A_i \cap A_j \big| = 0$$
And going back to the fact that $\frac{15}{6} = \frac{5}{2}$ each subject must contain at-least two common elements with another subset.
Is my proof correct? - I think I may be over thinking it...
Just apply the theorem you quote. The union on the left is no larger than $15$. The first sum on the right is $30$ and there are $10$ items in the second sum on the right. If each item is of size $1$ the right is $20$, a contradiction.