Apply Martin's axiom at a Kunen's exercise I need help in the following exercise from Chapter II of Kunen's first edition set theory book.


Hint 1: Consider $\mathbb{P}=\{\langle p,n\rangle \ : \ n\in\omega , dom(p)\subseteq X \mbox{ is finite and } p(x)\subseteq n \mbox{ for all } x\in dom(p)  \}$ with the ordering $\langle p,n \rangle \leq \langle q,m \rangle$ iff $m\leq n, dom(q)\subseteq dom(p), \ \forall x\in dom(q) (p(x)\cap m =q(x)),\ $ and $\forall x,y \in dom(q) (x<y \implies(p(x)\setminus p(y))\subseteq m)$.


Hint 2: Use the $\Delta$-system lema to prove that $\mathbb{P}$ has the countable chain condition property.

The $\Delta$-system lemma says that whenever we have an uncountable family of finite sets, we can find an uncountable sub-family $\mathcal{D}$ that forms a $\Delta$-system, i.e., there is a $r\in\mathcal{D}$ (the root) such that for any two elements $x,y\in\mathcal{D}$ we have $x\cap y=r$.
Well, of course we only have to construct smart $\kappa$ dense subsets of $\mathbb{P}$ and apply  martin's axiom in those denses and get a generic filter which will probably sinalize which $a_x$ we should define to finish the exercise. But I struggled at the very beggining. I could not define those dense subsets in a smart way... it is my first exercise in $MA$ issues so I don't have some ideas to follow. Could you help me?
Also, I couldn't use Hint 2 to prove that $\mathbb{P}$ is c.c.c.. On this, I supposed an uncountable antichain $\mathcal{A}$ and  tried to define $\mathcal{A}'=\{dom(p): \exists n(\langle p,n \rangle \in  \mathcal{A})\}$ to apply the $\Delta$-system lema and get any contradiction, but I didn't go further on that.
Thanks in advance.
 A: $\newcommand{\dom}{\operatorname{dom}}$
Revised and corrected 20 June 2021.
For $x\in X$ and $k\in\omega$ let
$$\begin{align*}
D_{x,k}=\{\langle p,n\rangle&\in\Bbb P:x\in\dom(p)\text{ and }|p(x)|\ge k\text{ and }\\
&\forall y\in\dom(p)\setminus\{x\}(x<y\to|p(y)\setminus p(x)|\ge k)\}\;.
\end{align*}$$
Let $\langle q,m\rangle\in\Bbb P$ be arbitrary. If $x\notin\dom(q)$, let $q'=q\cup\{\langle x,m\rangle\}$; otherwise let $q'=q$. Let
$$s=(m+k+1)\setminus(m+1)$$
and
$$t=(m+2k+1)\setminus(m+k+1)\,,$$
let $n=m+2k+1$, and define $p:\dom(q')\to n$ as follows:
$$p(y)=\begin{cases}
q'(y)\cup s,&\text{if }y=x\\
q'(y)\cup s\cup t,&\text{if }x<y\\
q'(y),&\text{otherwise.}
\end{cases}$$
Then $\langle p,n\rangle\le\langle q,m\rangle$, and $\langle p,n\rangle\in D_{x,k}$, so $D_{x,k}$ is dense in $\Bbb P$. For convenience let $D_x=\bigcup_{k\in\omega}D_{x,k}$ for each $x\in X$.
Let $G$ be a filter in $\Bbb P$ meeting each $D_{x,k}$. For $x\in X$ let
$$a_x=\bigcup_{\langle p,n\rangle\in G\cap D_x}p(x)\;$$
clearly $a_x\subseteq\omega$, and the fact that $G$ meets $D_{x,k}$ for each $k\in\omega$ ensures that $|a_x|=\omega$. Suppose that $x,y\in X$ and $x<y$; we want to show that $|a_x\setminus a_y|<\omega$.
If $a_x\setminus a_y\ne\varnothing$, let $\ell\in a_x\setminus a_y$; $G$ is a filter, so there is a $\langle p,n\rangle\in G\cap D_x\cap D_y$ such that $\ell\in p(x)$. Then for each $\langle q,m\rangle\in G$ such that $\langle q,m\rangle\le\langle p,n\rangle$ we have $q(x)\setminus q(y)\subseteq n$, so $a_x\setminus a_y\subseteq n$ and therefore $a_x\subseteq^*a_y$.
Moreover, for each $k\in\omega$ there is a $\langle p,n\rangle\in G\cap D_{x,k}\cap D_y$, so that $|p(y)\setminus p(x)|\ge k$. If $\langle q,m\rangle\in G$ with $\langle q,m\rangle\le\langle p,n\rangle$, then $\big(q(y)\setminus q(x)\big)\cap n=p(y)\setminus p(x)$, so $(a_y\setminus a_x)\cap n=p(y)\setminus p(x)$, and $|a_y\setminus a_x|\ge k$. Thus, $a_y\setminus a_x$ is infinite, and $a_x\subset^*a_y$.
To show that $\Bbb P$ is ccc, let $A\subseteq\Bbb P$ be uncountable; WLOG we may assume that there is an $n_0\in\omega$ such that $n=n_0$ for each $\langle p,n\rangle\in A$, and we may further assume that that $\{\dom(p):\langle p,n_0\rangle\in A\}$ is a $\Delta$-system with root $r$. And $n_0$ has only finitely many subsets, so we may assume that $p\upharpoonright r=q\upharpoonright r$ for all $p,q\in A$.
Now let $\langle p,n_0\rangle,\langle q,n_0\rangle\in A$, and let $s=p\cup q$; then $s:\dom(p)\cup\dom(q)\to n_0$, so $\langle s,n_0\rangle\in\Bbb P$, and it’s easy to check that $\langle s,n_0\rangle\le\langle p,n_0\rangle$ and $\langle s,n_0\rangle\le\langle q,n_0\rangle$, so $A$ is not an antichain, and $\Bbb P$ is ccc.
