Let $W$ be a 7-dimensional space. Prove that for any 5-dimensional subspaces $W_1,W_2,W_3 \subset W$ the intersection $W_1 \cap W_2 \cap W_3$ is non-zero.
Proof by contradiction. Suppose not, then $W_1$ and $W_2 \cap W_3$ are disjoint. Take a basis of each. Because they are disjoint you can combine these two bases to get a linearly independent set in $W$. This proves that:
$$\dim W_1 + \dim(W_2 \cap W_3) \leq \dim W$$
$$\dim(W_2 \cap W_3) \leq 2.$$
Is this possible if both $W_2$ and $W_3$ are $5$-dimensional? To see take a basis for $W_2 \cap W_3$ and then extend this to a basis for $W_2$ and a basis for $W_3$. These vectors are all linearly independent so there can't be more than $7$ of them. Count them up and see how many you actually get.