$\newcommand{\ms}{\mathscr}$Please give me only a small hint, or give me an easier problem to solve that would help me do this.
A sunflower is a family of sets (petals) for which ever pairwise intersection is the same.
I want to prove
Sunflower Lemma: Let $\ms{A}$ be a family of sets such that $|A|\le\ell$ for each $A\in\ms{A}$. If $|\ms{A}|>(p-1)^\ell\ell!$, then $\ms{A}$ contains a sunflower with $p$ petals.
(Thanks to Brian M Scott for the correct statement of the lemma)
I though about this a lot but I didn't get much. First I realized that a family of sets like {1,2,3,4,5}, {1,2,3,4}, {1,2,3}, {1,2}, {1} is never a sunflower.
I also found that if there is no sunflower with empty core (I named the pairwise intersection the core) then our sets are made out of at most $lp$ symbols.
It's also impossible for our sunflower to have full core (equal to the whole set) because there would only be one set in our family of sets.
I tried to think a bit about the probability of two sets having an intersection, but it didn't seem fruitful because it's so complex to try to generalize this to more sets.
I was thinking for a while about the largest possible set containing no sunflower. If it didn't have any empty set in it's pairwise intersection then out of every $p$ sets at least one pair must have a nonempty intersection. I tried to use pigeonhole principle to get 3 petals but this gave a too weak bound.
I am thinking about some way of doing the induction step for $\ell$ like for sets of size $\ell$ then $|\ms{A}|>N=(p-1)^\ell\ell!$ has n $p$-sunflower $\implies |\ms{A}|>N \cdot (p-1)(\ell+1)$ for sets of size $\ell+1$ has a $p$-sunflower. The is that if you allow the sets to be bigger, they're much more capable of avoiding being sunflowers. The reason for that might be related to my first observation. If I have $N \cdot (p-1)(\ell+1)$ then I was first thinking each of the $(p-1)(\ell+1)$ blocks of size $N$ have a sunflower in them but I don't actually have that so I need a new idea.
From Andres Caicedo hints I have the following:
The proof should work by induction on $\ell$, so suppose than any family of sets of size $\ell$ will contain a $p$-petalled sunflower if $|\ms{A}| > N$, we would like to find a bound on the cardinality of a family of sets of size $\ell+1$ needed for it to contain a $p$-petalled sunflower.
Suppose there are not $p$ pairwise disjoint sets, since if there were we would be done.
Let $\ms{F}$ be a maximal disjoint subfamily of $\ms{A}$, then every set from $\ms{A}$ has nonempty intersection with exactly one set of $\ms{F}$ (otherwise it should have been included in $\ms{F}$ since it's disjoint to everything in there). This allows us to partition $\ms{A}$ into families $\ms{A}_F$ for each $F \in \ms{F}$ and any sunflower must be completely contained in one of the families.
Since there are at most $p-1$ of the $\ms{A}_F$, we we would be able to get a bound for $\ms{A}$ and be done by pigeonholing if we could just figure out how large an $\ms{A}_F$ needs to be for there to be a $p$-petalled sunflower in it. This plan seems promising but I don't quite see how we could go from the idea of induction on $\ell$ to the idea of taking a maximal disjoint subfamily.
If any element was in more than $N$ of sets then we would have a sunflower by the induction hypothesis (just remove the element and we have a family of $N$ $\ell$ sets). So I asked: How many sets would we need to meet $F$ in order for an element to be contained in $N$ of them? The answer is $N|F|$ which is at most $N (\ell+1)$.
The pigeonholing argument gives us that we need $N \cdot (p-1)(\ell+1)$ sets in $\ms A$.