Is there more simple way to get a doughnut property?

The terms and my question appear from the Halbeisen's book Combinatorial set theory with a gentle introduction to forcing.

For subsets $a$, $b$ of $\omega$ such that $b-a$ is infinite, define a doughnut as $$[a,b]^\omega:= \{x\in [\omega]^\omega : a\subseteq x\subseteq b\}.$$ (Imagine the Venn diagram of $a$, $x$ and $b$ then you can guess why this set is called a doughnut.)

A collection $\mathcal{A}\subset [\omega]^\omega$ has a doughnut property if there are $a$ and $b$ such that either $[a,b]^\omega\subseteq \mathcal{A}$ or $[a,b]^\omega\cap \mathcal{A} = \varnothing$ hold. If previous conditions hold for $a=\varnothing$, we call $\mathcal{A}$ has the Ramsey property.

Under the axiom of choice we can find a collection $\mathcal{A}$ that has a doughnut property but not Ramsey property. Here is my attempt:

Define $\sim$ as $$a\sim b \iff a\triangle b\text{ finite}.$$ Then $\sim$ is an equivalence relation. For each $x\in [\omega]^\omega$ choose a representative $r_x$. (that is, choose $r_x$ such that $[x]=[r_x]$ for each equivalence class.) Take a maximal antichain $X\subseteq[\omega]^\omega$ which is not contain $\omega$.

Let define $\mathcal{A}\subseteq[\omega]^\omega$ as follow: for given $y\in[\omega]^\omega$,

1. if there is some $a\in X$ such that $a\subseteq y$, then $y\in\mathcal{A}$.

2. if not, $y\in\mathcal{A}$ if and only if $|y\triangle r_y|$ is even.

then for $a\in X$, $[a,\omega]^\omega\subseteq\mathcal{A}$. But for any $y\in [\omega]^\omega$, we can find some $a\in X$ such that $a$ and $y$ are comparable and

1. if $y\subseteq a$ then we can find some subsets of $y$ contained in $\mathcal{A}$ and not in $\mathcal{A}$.

2. if $a\subseteq y$, we can consider a set $b\subseteq a$ such that $a-b$ is infinite. (since $a$ is infinite) then $b$ has subsets lies on $\mathcal{A}$ and not on $\mathcal{A}$.

thus $[y]^\omega$ is neither a subset of $\mathcal{A}$ nor disjoint from $\mathcal{A}$.

My questions are: Is my construction valid? and is there more simple construction?

Sorry I don't have enough credits to comment, so I wrote down my construction here as an answer although I am not sure whether it is correct.

We can always find $r_{a}$ and $r_{b}$ such that $r_{b}-r_{a}$ is infinite.
Let $a = r_{a}$ and $b = r_{b}$.
$[a,b]^{\omega}=\{x\in[\omega]^{\omega}:r_{a}\subseteq x\subseteq r_{b}\}$.
$A=\{x\in[\omega]^{\omega}:|x\triangle r_{x}|\ is\ even\}\cup\{x\in[\omega]^{\omega}:r_{a}\subseteq x\subseteq r_{b}\}$.
For arbitrary $x\in[\omega]^{\omega}$, there are 2 situations:

1. if $|x\triangle r_{x}|$is even, then $x\in A$. We can choose an element $a\in x$ then $|x-\{a\}\triangle r_{x}|$ will be odd.
   If $x-\{a\}$ is not in $[a,b]^{\omega}$, then we are done. If so, because $r_{a}\subseteq x$, we can find two infinite subset of $r_{a}$ (also of $x$) $x_{1} $and $x_{2}$ such that $|x_{1}\triangle r_{a}|$is odd and $|x_{2}\triangle r_{a}|$is even, for example, take one element out of $r_{a}$, take two out of $r_{a}$.
2. if$|x\triangle r_{x}|$is odd, we can choose an element $a\in x$ then $|x-\{a\}\triangle r_{x}|$ will be even.