Solution
= Solution
Write $[A]^\omega$ for the infinite subsets of an infinite $A\subseteq\mathbb N$. A <Ramsey set of infinite subsets> is a family $E\subseteq[\mathbb N]^\omega$ such that each infinite ground set $A$ has an infinite subset $B$ with
$$
[B]^\omega\subseteq E\qquad\text{or}\qquad [B]^\omega\cap E=\varnothing.
$$
The convention asking only for one such homogeneous cone is weaker; the proof below establishes the displayed property and, in fact, the stronger <completely Ramsey set> property.
For a finite stem $s$ and an infinite reservoir $A$ lying above every element of $s$, put
$$
[s,A]=\{s\cup X:X\in[A]^\omega\}.
$$
These sets form the basic open <neighbourhoods> of the <Ellentuck topology>, or <star topology>. Here $s$ is an initial segment of every member of $[s,A]$. A star-<Borel set> means a member of the <sigma-algebra> generated by these <open sets>. A family is <completely Ramsey> if every $[s,A]$ has a homogeneous refinement $[s,B]$ with $B\subseteq A$ infinite and the same stem $s$. We prove this for every star-<Borel set>, beginning with <open sets>.
Fix a star-open family $O$ and a basic <neighbourhood> $[s,A]$. An infinite reservoir $T$ above a finite stem $t$ accepts $t$ if $[t,T]\subseteq O$. It rejects $t$ if no infinite subset of $T$ accepts $t$. A reservoir decides $t$ if it accepts or rejects it. Any reservoir has a refinement deciding a prescribed stem: take an accepting refinement if there is one, and otherwise the current reservoir rejects. Both kinds of decision persist under infinite thinning.
First refine $A$ to decide $s$. If the decision is acceptance, the required homogeneous <neighbourhood> has already been found. Suppose it is rejection. We use <deciding all finite stems by fusion> to select
$$
b_1<b_2<\cdots
$$
from nested infinite reservoirs. After selecting $b_n$, successively thin the remaining reservoir to decide every stem $s\cup a$ with $a\subseteq\{b_1,\ldots,b_n\}$. There are only finitely many such $a$, so every stage can be completed. Choose $b_{n+1}$ from that thinned reservoir and repeat. Let $B=\{b_1,b_2,\ldots\}$. For every finite $a\subseteq B$, the tail
$$
B/a=\{b\in B:b>\max a\}
$$
decides $s\cup a$; for $a=\varnothing$, use all of $B$. Indeed, the stem was decided at the stage when its largest point was selected, and its final tail is a subset of the reservoir retained at that stage. The empty stem remains rejected by the initial decision.
The crucial finite-obstruction fact is that a rejected stem has only finitely many accepting one-point extensions. More precisely, suppose $B/a$ rejects $s\cup a$. If infinitely many $b\in B/a$ had $B/b$ accepting $s\cup a\cup\{b\}$, collect those $b$ into an infinite $C$. Any $X\in[C]^\omega$ has a least point $b$, and its remaining tail lies in $B/b$. Acceptance at $b$ would then imply $s\cup a\cup X\in O$. Thus $C$ would accept $s\cup a$, contradicting rejection. This proves the <finitely many accepting extensions of a rejected stem> assertion with the reservoirs specified.
Thin $B$ a second time. Having selected a finite prefix of $C$, maintain that every stem $s\cup a$ with $a$ a subset of that prefix is rejected relative to its tail in $B$. Each such stem has finitely many forbidden next points leading to acceptance. Choose the next point of $C$ above the prefix and outside the union of these finitely many finite forbidden sets. The deciding property of $B$ then makes every newly created stem rejected. Starting with the rejected empty extension gives an infinite $C\subseteq B$ all of whose finite extensions of $s$ are rejected.
We claim $[s,C]\cap O=\varnothing$. Otherwise take $Y\in[s,C]\cap O$. Star-openness supplies a basic <neighbourhood> inside $O$ containing $Y$. Extend its finite stem along $Y$ until it contains $s$; the new stem has the form $s\cup a$ for a finite initial segment $a$ of $Y\setminus s$. The remaining infinite tail of $Y$ is a subset of $B/a$ and accepts that stem, contradicting rejection. This proves the <fusion proof for open Ellentuck sets>: \b[every star-<open set> is completely Ramsey].
Next call $N$ <completely Ramsey-null> if every $[s,A]$ has a same-stem refinement disjoint from $N$. Every star-<nowhere dense set> is completely Ramsey-null. Indeed, the complement $D$ of its star-closure is star-open and <dense>. Apply the just-proved open-set result inside $[s,A]$. The homogeneous alternative disjoint from $D$ would be a nonempty <open set> contained in the closure of a nowhere-dense set, which is impossible. Hence some $[s,B]$ lies entirely in $D$ and avoids $N$.
Countable unions of completely Ramsey-null sets are again completely Ramsey-null. Here is the full <Ellentuck meagre-set fusion lemma>. Given $N_1,N_2,\ldots$ and $[s,A]$, construct selected prefixes $P_j=\{b_1,\ldots,b_j\}$ and nested unused infinite reservoirs. At stage $j$, refine the reservoir, still lying above $\max P_j$, to arrange
$$
[s\cup a,T_j]\cap N_i=\varnothing\qquad
\text{for every }a\subseteq P_j\text{ and every }1\le i\le j+1.
$$
Each constraint can be imposed by complete Ramsey-nullness, with all earlier constraints preserved by thinning; there are finitely many constraints. Choose $b_{j+1}\in T_j$ and retain the tail above it for the next stage. Start with $P_0=\varnothing$ and the reservoir $A$.
Let $B=\{b_1,b_2,\ldots\}$. If $Y\in[s,B]$ and $i$ is fixed, take any $j\ge i-1$ and put $a=(Y\setminus s)\cap P_j$. Every remaining point of $Y\setminus s$ lies in $T_j$, so $Y\in[s\cup a,T_j]$ and $Y\notin N_i$. Therefore
$$
[s,B]\cap\bigcup_{i=1}^\infty N_i=\varnothing.
$$
This construction handles all subsets of a selected prefix, not merely the full prefix: members of $[s,B]$ may skip arbitrarily many selected points. Since a <meagre set> is a countable union of nowhere-dense sets, \b[every star-<meagre set> is completely Ramsey-null].
It remains to pass from star-<open sets> to star-<Borel sets>. Let $\mathcal C$ consist of the families $E$ expressible as
$$
E\triangle O\subseteq M,
$$
where $O$ is star-open and $M$ is star-meagre. We verify directly that $\mathcal C$ is a <sigma-algebra>. It contains every <open set>. Countable unions remain in $\mathcal C$, because
$$
\left(\bigcup_i E_i\right)\triangle\left(\bigcup_i O_i\right)
\subseteq\bigcup_i(E_i\triangle O_i),
$$
and a countable union of <meagre sets> is meagre. For complements, replace $O^c$ by the <open set> $\operatorname{int}(O^c)$. The omitted part is $\overline O\setminus O$, the boundary of the <open set> $O$, which is closed and <nowhere dense>: a nonempty open subset of $\overline O$ must meet $O$, so it cannot lie in this boundary. Thus
$$
E^c\triangle\operatorname{int}(O^c)\subseteq M\cup(\overline O\setminus O),
$$
a <meagre set>. Subsets of <meagre sets> are meagre as well, since taking a subset preserves nowhere denseness. These observations prove the claimed <sigma-algebra> property. Hence every star-<Borel set> belongs to $\mathcal C$.
Finally, for $E\triangle O\subseteq M$ and any $[s,A]$, first use complete Ramsey-nullness to obtain $[s,B]$ avoiding $M$. On that neighbourhood $E$ and $O$ coincide. Apply the open-set result to $O$ within $[s,B]$ to get a homogeneous $[s,C]$. Therefore
$$
\boxed{\text{Every star-Borel family is completely Ramsey, and hence Ramsey.}}
$$
Taking $s=\varnothing$ gives the requested homogeneous infinite-subset cone inside any given infinite ground set. Every closure, interior, <open set> and meagreness used in this proof refers to the <Ellentuck topology>; substituting the coarser ordinary topology at an intermediate step would change the assertion.