Recently, I asked a question about whether we can well-order $\mathbb R$ over $\sf ZF$ using Sierpinski’s result ($2^{\aleph_0} \nrightarrow (\aleph_1, \aleph_1)^2)$.
As pointed out in the comments, the statement holds trivially, for example, in Solovay’s model since there are no subsets of $\mathbb R$ with cardinality $\aleph_1$ in Solovay’s model. Since $\mathbb R$ can’t be well-ordered in Solovay’s model, the result does not imply the fact that $\mathbb R$ can be well-ordered.
Also, if $\mathbb R$ is a countable union of (WLOG disjoint) sets, say, $\mathbb R = \bigsqcup_{i \in \mathbb N} A_i$, where each $A_i$ is countable, then we can colour $\{x,y\}$ blue if $x$ and $y$ belong to the same $A_i$, red otherwise. It is clear that with respect to this colouring, $\mathbb R$ can’t have a monochromatic subset of the desired cardinality. So, the statement holds in any model in which $\mathbb R$ is a countable union of countable sets, so, for example, it does not imply countable choice.
After seeing these, I think this statement can be proved without choice at all, so I wanted to ask whether there is a proof of this fact which does not use choice of any form, thank you.
The following theorem is due to Galvin: (see e.g., here )
So if a model satisfies $\aleph_1\leq 2^{\aleph_0}$ + every set of reals has the Baire property + Galvin's theorem (which I believe uses ZF+DC), then that model will also have $2^{\aleph_0}\to(\aleph_1, \aleph_1)^2$.
In Shelah's model ($\mathcal{M}18$ in the Howard&Rubin book), every set has the BP and DC holds. Initially I wasn't sure whether $\aleph_1\leq 2^{\aleph_0}$ holds in that model. But as @ElliotGlazer kindly pointed out, since Shelah's model doesn't require an inaccessible to build, we can assume that there are no inaccessibles in $L$. This will imply $\aleph_1\leq 2^{\aleph_0}$ (because DC and $\aleph_1\not\leq 2^{\aleph_0}$ would imply that $\aleph_1$ is inaccessible in $L$).
So Sierpinski's coloring result does not hold in Shelah's model.