= Solution
Write $[M]^\omega$ for the <set> of all infinite <subsets> of $M$. A family $Y\subseteq[\mathbb N]^\omega$ is a <Ramsey family in the homogeneous-cone sense> if there is an infinite $M$ such that either $[M]^\omega\subseteq Y$ or $[M]^\omega\cap Y=\varnothing$. Thus membership in $Y$ is constant on an infinite homogeneous cone. We will prove the stronger version for open families: such a cone exists inside every prescribed infinite ground <set>, as in the definition of a <Ramsey set of infinite subsets>.
The symbols in the question refer to the <ordinary topology on infinite subsets> $\tau$ and the <Ellentuck topology> $*$. For a <finite set> $s$, let
$$
[s]=\{X\in[\mathbb N]^\omega:s\text{ is the initial segment of }X\}.
$$
These <sets> form a <topological basis> for $\tau$, equivalently the <subspace topology> inherited from the <product topology> on $\{0,1\}^{\mathbb N}$. For an infinite $A$ with $\max s<\min A$, the <sets>
$$
[s,A]=\{s\cup B:B\in[A]^\omega\}
$$
form a <topological basis> for $*$; take $\max\varnothing=0$. In particular $[s]=[s,\{n:n>\max s\}]$, so $*$ is finer than $\tau$.
For a non-Ramsey example use the <finite-symmetric-difference parity colouring>. On $[\mathbb N]^\omega$, define the <equivalence relation> $X\sim Z$ if their <symmetric difference> $X\mathbin\triangle Z$ is finite. By the <axiom of choice>, select a representative $R$ in every <equivalence class>. Give $X$ the colour
$$
c(X)=|X\mathbin\triangle R_X|\pmod2,
$$
where $R_X$ is its class representative. Deleting one point stays in the same class and toggles the parity. Thus for every infinite $M$, the <sets> $M$ and $M\setminus\{\min M\}$ are both in $[M]^\omega$ but have opposite colours. Either <colour class>, regarded as a family $Y$, is therefore \b[not Ramsey].
Now let $U$ be $\tau$-open. We prove that for every infinite $A_0$ there is an infinite $C\subseteq A_0$ with $[C]^\omega\subseteq U$ or $[C]^\omega\cap U=\varnothing$. The proof uses <acceptance and rejection of finite stems>, which we define explicitly. For a <finite stem> $s$ and an infinite tail $A$ above $\max s$, say that $A$ accepts $s$ if $[s,A]\subseteq U$, and rejects $s$ if no infinite <subset> of $A$ accepts $s$. Both properties persist on taking infinite <subsets>. Every tail has an infinite <subset> that decides $s$: take an accepting <subset> if one exists, and otherwise it already rejects $s$.
First thin $A_0$ to decide the empty stem. Then construct $b_1<b_2<\cdots$ and nested infinite tails. At step $n$, choose $b_n$ from the current tail, discard its points at most $b_n$, and successively thin what remains to decide every stem $s\subseteq\{b_1,\ldots,b_n\}$. There are only finitely many such stems, so an infinite tail remains. Let $B=\{b_1,b_2,\ldots\}$ and write $B/s=\{b\in B:b>\max s\}$. For every finite $s\subseteq B$, the tail $B/s$ decides $s$, since it lies in the tail obtained at the stage when the largest point of $s$ was selected. The empty stem is also decided. This is <deciding all finite stems by fusion>.
If $B$ accepts the empty stem, then $[B]^\omega\subseteq U$, as required. Otherwise $B$ rejects it. We need the <finitely many accepting extensions of a rejected stem> observation. Suppose a tail $B/s$ rejects $s$. Only finitely many $a\in B/s$ can have $B/a$ accepting $s\cup\{a\}$. Indeed, if infinitely many such $a$ formed a <set> $D$, every member of $[s,D]$ would have some first new point $a$ and all later points in $B/a$, hence would lie in $U$. This would make $D$ an accepting <subset> of $B/s$, contradicting rejection.
Choose $c_1<c_2<\cdots$ in $B$ recursively so that for every finite $s\subseteq\{c_1,\ldots,c_n\}$, the tail $B/s$ rejects $s$. The empty stem satisfies this initially. At the next step, each of the finitely many previous stems has only finitely many accepting one-point extensions by the observation. Choose $c_{n+1}$ larger than all previous points and outside all those finite exceptional <sets>. The first fusion ensured that the tail for every new stem decides it; since it does not accept it, it rejects it. This maintains the recursion.
Put $C=\{c_1,c_2,\ldots\}$. If some $X\in[C]^\omega$ belonged to $U$, ordinary openness would give a finite initial segment $s$ of $X$ with $[s]\subseteq U$. In particular $[s,C/s]\subseteq U$. But $C/s$ is an infinite <subset> of $B/s$, which rejects $s$, a contradiction. Hence $[C]^\omega\cap U=\varnothing$. We have proved
$$
\boxed{\text{Every }\tau\text{-open family is Ramsey, even inside every infinite ground set}.}
$$
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