If we define $\pi: [\mathbb N]^2 \rightarrow \{0,1\}$ randomly (say each $\pi(a,b)$ is determined by a coin flip), then almost surely there is no set $S$ with $d(S) > 0$ that is Ramsey for $\pi$. In fact, it is almost surely true that every $S$ with $d(S) > 0$ contains an induced isomorphic copy of the randomly colored graph.

Even more: for a random coloring $\pi$ of $[\mathbb N]^2$, there is almost surely no set $S$ with $\sum_{n \in S \setminus \{0\}} \frac{1}{n} = \infty$ that is Ramsey for $\pi$. In fact, it is almost surely true that every such $S$ contains an induced copy of every coloring of every finite graph. (But in this case, "finite" cannot be improved to "infinite" as above.)

These results can be found in Section 2 of my paper "Which subsets of the infinite random graph look random?" (Mathematical Logic Quarterly 64 (2018), pp. 478-486), available here.

If we define $\pi: [\mathbb N]^2 \rightarrow \{0,1\}$ randomly (say each $\pi(a,b)$ is determined by a coin flip), then almost surely there is no set $S$ with $d(S) > 0$ that is Ramsey for $\pi$. In fact, it is almost surely true that every $S$ with $d(S) > 0$ contains an induced isomorphic copy of the randomly colored infinite graph.

Even more: for a random coloring $\pi$ of $[\mathbb N]^2$, there is almost surely no set $S$ with $\sum_{n \in S \setminus \{0\}} \frac{1}{n} = \infty$ that is Ramsey for $\pi$. In fact, it is almost surely true that every such $S$ contains an induced copy of every coloring of every finite graph. (But in this case, "finite" cannot be improved to "infinite" as above.)

These results can be found in Section 2 of my paper "Which subsets of the infinite random graph look random?" (Mathematical Logic Quarterly 64 (2018), pp. 478-486), available here.