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 $y\in[\omega]^\omega$,
  
  
  if there is some $a\in X$ such that $a\subseteq y$, then $y\in\mathcal{A}$.
  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
  
  
  if $y\subseteq a$ then we can find some subsets of $y$ contained in $\mathcal{A}$ and not in $\mathcal{A}$.
  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?
I think your construction works fine, adding the additional condition on the maximal antichain $X$: it should contain an infinite set $a$, whose complement is also infinite (that such $X$ exists can be quickly established by Zorn's lemma). This condition is needed for the doughnut $[a, \omega]^{\omega} \subseteq \mathcal{A}$ (the definition of the doughnut requires that $\omega \setminus a$ is infinite).
I think your construction works fine, adding the additional condition on the maximal antichain $X$: it should contain an infinite set $a$, whose complement is also infinite (that such $X$ exists can be quickly established by Zorn's lemma). This condition is needed for the doughnut $[a, \omega]^{\omega} \subseteq \mathcal{A}$ (the definition of the doughnut requires that $\omega \setminus a$ is infinite).