A ccc tree poset whose square is not ccc

F1 identifies every poset antichain in $P$ with a tree antichain, which is countable by F4. Thus $P$ is ccc. Its root is greatest in the reverse order, since every node has that unique root below it by F5. Therefore F2 applies to its two-factor product.

Every $t\in T$ has two distinct immediate successors, since $\mathrm{ht}(t)+1<{\omega }_1$; choose an ordered pair $(t^0,t^1)$ simultaneously for all $t$ using F3 and A1. Each has height $\mathrm{ht}(t)+1$: a larger height would give an intermediate predecessor by F5. Define $e(t)=(t^0,t^1)$. This map is injective: equality of its first coordinates forces equal parent heights and then identical predecessors at that height by F5. The tree is uncountable, because its height map is onto ${\omega }_1$ (use height and F5 for nonempty levels) and a countable set cannot have uncountable image. Hence $e[T]$ is uncountable.

For incomparable parents $t,u$, the nodes $t^0,u^0$ cannot be comparable: a comparison would give a common extension of $t,u$, forcing them comparable by F5. Thus these product pairs are incompatible by F1 and F2. For comparable distinct parents, interchange their names if necessary so that $t{<}_{T}u$. Suppose both coordinates of $e(t),e(u)$ were compatible. F1 and heights then give $t^i{\le }_T{u}^i$ for $i=0,1$. Both $t^i$ and $u$ are below $u^i$, so are comparable by F5. Since $\mathrm{ht}(t^i)=\mathrm{ht}(t)+1\le \mathrm{ht}(u)$, this forces $t^i{\le }_{T}u$. Unique predecessors at that height (or equality when heights coincide) would give $t^0=t^1$, contrary to their choice. Hence at least one coordinate is incompatible, and F2 makes the product pairs incompatible. Thus $e[T]$ is an uncountable product antichain, proving the square is not ccc.