Ccc and proper are not equivalent

Let $P=\mathrm{Fn}({\omega }_1,2,<{\omega }_1),$ ordered by reverse inclusion. By F2 its conditions are the countable partial functions from ${\omega }_1$ to $2$, and the empty function is its greatest condition. In particular, $P$ is nonempty. (It is not being identified with $\mathrm{Col}({\omega }_1,2)$, whose displayed parameters would violate that definition's requirement $\kappa \le \lambda$.)

For every $\alpha <{\omega }_1$, define $p_{\alpha }\in P$ on $\alpha +1$ by $p_{\alpha }(\xi )=\{\begin{array}{ll}0,& \xi <\alpha ,\\ 1,& \xi =\alpha \mathrm{.}\end{array}$ The domain is countable because $\alpha$ is a countable ordinal. If $\alpha <\beta <{\omega }_1$, then $p_{\alpha }(\alpha )=1$ whereas $p_{\beta }(\alpha )=0$. No function can extend both, so $p_{\alpha }$ and $p_{\beta }$ are incompatible. Consequently $\{p_{\alpha }:\alpha <{\omega }_1\}$ is an uncountable antichain, and F3 shows that $P$ is not ccc. Notice that $p_0=\{(0,1)\}$, so the zero endpoint also obeys the displayed definition.

Suppose $⟨q_n:n<\omega ⟩$ is descending in $P$. Reverse inclusion means $q_n\subseteq q_{n+1}$, so $q={\bigcup }_{n<\omega }{q}_n$ is a function extending every $q_n$. Each $\mathrm{dom}(q_n)$ is countable, and F4 with A1 makes their union countable. Hence $q\in P$ and $q\le q_n$ for every $n$. Thus $P$ is countably closed. Constant sequences, including the constant empty-condition sequence, are covered by the same union calculation.

By F1, the countably closed forcing $P$ is proper.

The forward implication, ccc implies proper, is true by F1. Steps 3.1 and 1.2 give one proper forcing that is not ccc, so the reverse implication and therefore the advertised equivalence are false. AC is spent only through the countable-union assertion in step 2.1; no generic filter or further choice is used in the antichain witness.