[proofplan] We prove the contrapositive obstruction directly: an uncountable antichain would be an uncountable subset of $\mathbb{P}$, so the Knaster property would produce an uncountable subfamily whose distinct elements are pairwise compatible. But an antichain is precisely a subset whose distinct elements are pairwise incompatible. The same two distinct elements would therefore be both compatible and incompatible, a contradiction. [/proofplan] [step:Assume an uncountable antichain exists] Suppose, toward a contradiction, that $\mathbb{P}$ does not satisfy the countable chain condition. Then there exists an uncountable subset $A \subset \mathbb{P}$ such that $A$ is an antichain: for all distinct $p, q \in A$, the elements $p$ and $q$ are incompatible in $\mathbb{P}$. [/step] [step:Apply the Knaster property to the antichain] Since $A \subset \mathbb{P}$ is uncountable and $\mathbb{P}$ has the Knaster property, there exists an uncountable subset $B \subset A$ such that for all distinct $p, q \in B$, the elements $p$ and $q$ are compatible in $\mathbb{P}$. [guided] The Knaster property is applied exactly to the uncountable set $A$. Its hypothesis requires only that $A$ be an uncountable subset of $\mathbb{P}$, which holds by the preceding step. Therefore it gives an uncountable subset $B \subset A$ with the following pairwise compatibility property: whenever $p, q \in B$ and $p \neq q$, the elements $p$ and $q$ are compatible in $\mathbb{P}$. This is the decisive point: the Knaster property turns every uncountable subset into one containing many mutually compatible conditions. We apply it to a set that was assumed to be an antichain, so the conclusion will contradict the defining incompatibility of antichains. [/guided] [/step] [step:Contradict the defining incompatibility of an antichain] Because $B$ is uncountable, there exist distinct elements $p, q \in B$. Since $B \subset A$ and $A$ is an antichain, the elements $p$ and $q$ are incompatible in $\mathbb{P}$. Since $p$ and $q$ are distinct elements of $B$, the defining property of $B$ gives that $p$ and $q$ are compatible in $\mathbb{P}$. This contradiction shows that no uncountable antichain in $\mathbb{P}$ exists. Hence every antichain in $\mathbb{P}$ is countable, so $\mathbb{P}$ satisfies the countable chain condition. [/step]