Ccc posets are proper by maximal antichains

Fix a dense set $D\subseteq P$ with $D\in M$. By elementarity, inside $M$ choose a maximal antichain $A\subseteq D$. It is also maximal in $P$: maximality is the first-order assertion that every $r\in P$ is compatible with some $a\in A$, and all witnesses to compatibility are conditions in the ambient $H_{\theta }$. Since $P$ is ccc, $A$ is countable in the universe. Elementarity then puts in $M$ a surjection $e:\omega \to A$ (or a finite enumeration), and every $n<\omega$ belongs to $M$; hence $A\subseteq M$.

Let $r\le p$ be arbitrary. Maximality of $A$ gives $a\in A$ compatible with $r$. By step 1.1, $a\in A\subseteq D\cap M$. Thus $D\cap M$ is predense below $p$. Since this holds for every dense $D\in M$, $p$ is $(M,P)$-generic; the reflexive inequality $p\le p$ makes it a master below the original $p$.

The calculation works unchanged when $P$, $A$, or $D$ is finite. A dense subset of the stipulated nonempty $P$ cannot be empty, and for a one-condition order its unique condition is the required antichain member and master. No stronger condition than $p$ was constructed: ccc makes the starting condition itself sufficient.