Solution
= Solution
Identify $\mathbb N^{(\omega)}$ with the <space of infinite subsets of the natural numbers>. A family $E$ has an infinite <homogeneous set for a colouring> if some infinite $M$ satisfies
$$
\boxed{[M]^\omega\subseteq E\quad\text{or}\quad[M]^\omega\cap E=\varnothing.}
$$
This is the <Ramsey family in the homogeneous-cone sense> formulation. The relative <Ramsey set of infinite subsets> formulation asks for such an $M$ inside every prescribed infinite ground set. We prove the stronger <completely Ramsey> assertion, keeping an arbitrary <finite stem> fixed.
For a <finite stem> $s$ and an infinite tail $A$ with $\max s<\min A$ (taking $\max\varnothing=0$), write
$$
[s,A]=\{s\cup B:B\in[A]^\omega\}.
$$
These neighbourhoods generate the <Ellentuck topology>, or <star topology>. A star-<Borel set> belongs to the <sigma-algebra> generated by the <open sets> of this topology. It is <completely Ramsey> if every $[s,A]$ has an infinite $B\subseteq A$ with $[s,B]\subseteq E$ or $[s,B]\cap E=\varnothing$. Taking $s=\varnothing$ gives both preceding Ramsey formulations.
We first prove the <fusion proof for open Ellentuck sets>. Fix a star-<open set> $U$ and a starting neighbourhood $[s,A]$. A reservoir $C$ above a finite extension $s\cup t$ accepts $t$ when $[s\cup t,C]\subseteq U$. It rejects $t$ when no infinite subreservoir accepts it. Every infinite reservoir has a refinement deciding $t$: if an accepting subreservoir exists choose one, and otherwise the reservoir already rejects. Both acceptance and rejection persist under further infinite thinning. These are <acceptance and rejection of finite stems>.
First refine $A$ to decide the empty extension. Choose $d_0$ from the refined reservoir and thin the remaining tail to decide every subset of $\{d_0\}$. After choosing $d_0,\ldots,d_j$, thin the unused infinite tail to decide every subset of this finite prefix before choosing $d_{j+1}$. Only finitely many decisions occur at each stage. Let $D=\{d_0<d_1<\cdots\}$. For each finite $t\subseteq D$, its tail
$$
D_t=\{d\in D:d>\max t\}
$$
(with $D_\varnothing=D$) is contained in the reservoir chosen when the largest point of $t$ was selected. Thus $D_t$ decides $t$. This is <deciding all finite stems by fusion>.
If $D$ accepts the empty extension, $[s,D]\subseteq U$ and we are done. Otherwise it rejects it. Call a finite $t\subseteq D$ accepted or rejected according to the decision of $D_t$. For each rejected $t$, the set
$$
G_t=\{a\in D_t:t\cup\{a\}\text{ is accepted}\}
$$
is finite. Indeed, if $G_t$ were infinite, every $X\in[G_t]^\omega$ would have a least element $a$ with $t\cup\{a\}$ accepted, and its remaining tail lies in $D_{t\cup\{a\}}$. Hence $s\cup t\cup X\in U$ for every such $X$, making $G_t$ an accepting subreservoir for $t$, a contradiction. This proves the <finitely many accepting extensions of a rejected stem> fact.
Build $B=\{b_0<b_1<\cdots\}\subseteq D$ by choosing each new point outside all the finitely many sets $G_t$ for subsets $t$ of the already selected prefix. Begin with the rejected empty extension. Inductively every finite subset of the new prefix is rejected: the old subsets stay rejected, and each new subset is a rejected $t$ extended by a point outside $G_t$, whose decision must therefore be rejection. Consequently every finite $t\subseteq B$ is rejected.
If a point $X\in[s,B]$ belonged to $U$, star-openness would give a basic neighbourhood $[u,C]\subseteq U$ containing it. Extend $u$ along the increasing enumeration of $X$, if necessary, until its stem is $s\cup t$ for a finite $t\subseteq B$. The infinite tail of $X$ above that stem then provides an accepting subreservoir for $t$, contradicting rejection. Therefore $[s,B]\cap U=\varnothing$. We have proved that \b[every star-open set is completely Ramsey].
A family $N$ is <completely Ramsey-null> when every $[s,A]$ has a stem-preserving refinement disjoint from $N$. Every star-<nowhere dense set> has this property. Apply the open-set result to the dense open complement of its star-<closure>. The disjoint alternative would put an entire basic neighbourhood inside a <closure> with empty <interior>, which is impossible. The other alternative gives the required avoidance.
We also need the full <Ellentuck meagre-set fusion lemma>, not a diagonal argument checking only one prefix. Let $N_0,N_1,\ldots$ be <completely Ramsey-null>, and start inside $[s,A]$. At stage $j$, suppose $b_0,\ldots,b_{j-1}$ have been chosen. Refine the remaining infinite tail successively so that
$$
[s\cup t,C_j]\cap N_j=\varnothing
\quad\text{for every }t\subseteq\{b_0,\ldots,b_{j-1}\}.
$$
The null property permits each refinement with its stem unchanged; previous avoidance survives thinning. There are finitely many $t$ at this stage. Choose $b_j\in C_j$ and continue above it. Let $B=\{b_j:j\ge0\}$.
For any infinite $Y\subseteq B$ and any fixed $j$, put $t=Y\cap\{b_0,\ldots,b_{j-1}\}$. Its remaining points form an infinite subset of $C_j$, so $s\cup Y\in[s\cup t,C_j]$ and hence $s\cup Y\notin N_j$. This works for every $j$, proving
$$
[s,B]\cap\bigcup_{j=0}^\infty N_j=\varnothing.
$$
Thus countable unions of <completely Ramsey-null> sets are null. Subsets of null sets are null as well. In particular every star-<meagre set>, being a countable union of star-nowhere-dense sets, is completely Ramsey-null.
Finally let $\mathcal B$ consist of families $E$ such that $E\triangle U$ is contained in a star-meagre set for some star-open $U$. This is the <Baire property in the Ellentuck topology>. We verify that it forms a <sigma-algebra>. <Open sets> belong with empty error. For a countable union, use $U=\bigcup_j U_j$; its error is contained in the countable union of the meagre errors. For complements, replace $U^c$ by its <interior>. The discrepancy is
$$
U^c\setminus\operatorname{int}(U^c)=\overline U\setminus U,
$$
a closed nowhere-dense set: every <open set> meeting $\overline U$ also meets $U$, so none can lie wholly in this boundary. Adding it to the original meagre error handles complementation. Consequently every star-<Borel set> belongs to $\mathcal B$.
For such an $E$, start with any $[s,A]$. First apply null avoidance to its meagre error to find $[s,B]$ on which $E$ agrees with $U$. Then apply open-set homogeneity to refine it to $[s,C]$ wholly inside or outside $U$, and therefore wholly inside or outside $E$. This is the <Baire-property reduction for completely Ramsey families>. We conclude
$$
\boxed{\text{Every star-Borel family is completely Ramsey, and hence Ramsey.}}
$$
All closures, interiors, openness and meagreness in this argument refer to the <star topology>; replacing it midway by the ordinary finite-prefix topology would not justify the null-avoidance step.