Let $\mathbb P$ be a countably closed poset in which every condition has two incompatible extensions. Show that $\mathbb P$ does not satisfy the countable chain condition.
I tried direct proof.
I know that if $\mathbb P$ satisfies the ccc, every antichain in $\mathbb P$ is countable. Thus in order to show that $\mathbb P$ does not satisfy the countable chain condition, I need to show that there is at least one antichain $A \subseteq \mathbb P$ such that $A$ is uncountable. But I get stuck here as I want an $A$ to be uncountable but $A \subseteq \mathbb P$ and $\mathbb P$ is countable.
So I tried to prove by contradiction.
Suppose $\mathbb P$ is ccc. Then every antichain $A \subseteq \mathbb P$ is countable. Consider an antichain $A$ such that every condition has two incompatible extensions. I get stuck here as by definition an antichain $A$ has pairwise incompatible members. So I cant get the contradiction.
Help is much appreciated.
Here's a different direct proof, I'm not sure how to fix your arguments.
Start with a condition $p_{\langle\rangle}$, now define by recursion, for every finite binary sequence $s$, if $p_s$ is defined, then $p_{s^\frown 0}$ and $p_{s^\frown 1}$ are incompatible extensions of $p_s$. By $\sigma$-closure, we get that for every infinite binary sequence, $f$, we define $p_f$ to be some extension of the sequence $\{p_{f\restriction n}\mid n<\omega\}$.
Now prove that if $f\neq g$, then $p_f$ is incompatible with $p_g$. This means that we have an uncountable antichain.
What the above shows you is in fact more: if $\Bbb P$ is $\kappa$-closed, then it does not have $2^\lambda$-c.c., where $\lambda$ is the smallest such that $\kappa\leq 2^\lambda$.