Sunflowers in $\omega$ consisting of infinite sets If $X\neq\emptyset$ is a set, then ${\cal S}\subseteq {\cal P}(X)$ with ${\cal S}\neq \emptyset$ is said to be a sunflower if there is $K\subseteq X$ such that whenever $A\neq B\in{\cal S}$ ten $A\cap B = K$. ($K$ is sometimes said to be the kernel of ${\cal S}$, and it is allowed that $K = \emptyset$, in which case ${\cal S}$ is a collection of pairwise disjoint sets).
Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. There are examples of countable collections ${\cal K} \subseteq [\omega]^\omega$ such that for every sunflower ${\cal S}\subseteq {\cal K}$ we have $|{\cal S}| \leq 2$. (At the end of this post I construct such a ${\cal K}$.)
Question. Is there an uncountable collection ${\cal L}\subseteq [\omega]^\omega$ such that every sunflower ${\cal S}\subseteq {\cal L}$ has cardinality at most $2$?
Construction for ${\cal K}$. For every $n\in \omega$, let $p_n$ be the $n$th prime, so $p_0=2, p_1=3$, etc. For $n\in\omega$ set $$W_n = \{0, \ldots, p_n\} \cup \{p^n: n\in \omega\},$$ and let ${\cal K}=\{W_n: n\in \omega\}$.
 A: There exist uncountable chains in the poset $([\omega]^\omega,\subseteq)$ (e.g. by bijecting $\omega$ with $\mathbb Q$ and taking all downwards-closed sets). Let $L$ be such a chain. The only sunflowers in $L$ have two elements, since if $A,B,C\in L$ and $A\subsetneq B\subseteq C$, then $A\cap B=A\neq B=B\cap C$.
A: Let $C$ be the set of finite sequences of zeros and ones. I'll exhibit uncountably many infinite subsets of $C$ such that no three form a sunflower. Since $C$ is countably infinite, the example can be copied from $[C]^\omega$ to $[\omega]^\omega$.
For each infinite sequence $x$ of zeros and ones, let $A_x\subset C$ be the set of finite initial segments of $x$. Then, for any $x\neq y$, $A_x\cap A_y$ is determined by the longest common initial segment $s$ of $x$ and $y$, so the next terms in $x$ and in $y$ after $s$ are a 0 and a 1. For any sunflower $S\subset 
\{A_x:x:\omega\to 2\}$, say with kernel $s$, all its members would have different next terms after $s$. Since there are only two possible next terms, the sunflower has cardinality at most 2.
