It's easy to prove that, if $\mathbb{R}$ is well-orderable, then there is a 2-coloring of pairs of reals with no uncountable homogeneous set, i.e., there is an $m: [\mathbb{R}]^2\rightarrow 2$ such that for all uncountable sets $U\subseteq\mathbb{R}$, there are $x, y, z\in U$ such that $m(\lbrace x, y\rbrace)\not=m(\lbrace x, z\rbrace)$.

Proof: Let $<_w$ be a well-ordering of $\mathbb{R}$, and let $m(\lbrace x, y\rbrace)=0$ if $x< y\iff x<_w y$ and $m(\lbrace x, y\rbrace)=1$ if $x< y\iff y<_w x$. Then a homogeneous set for $m$ yields a well-ordered increasing or decreasing sequence of reals with the same cardinality. But there is no uncountable increasing or decreasing sequence of reals, since the reals have a countable dense subset. So we are done.

My question is, What happens if we don't assume that $\mathbb{R}$ is well-ordered? That is:

Question 1: Does there exist a model of $ZF$ in which $\mathbb{R}$ is not well-orderable but there is a 2-coloring of $\mathbb{R}$ with no homogeneous set of the same cardinality as $\mathbb{R}$?

A somewhat related question has to do with the complexity of colorings without nice homogeneous sets:

Question 2: Does every Borel 2-coloring of pairs of reals have a Borel homogeneous set with the same cardinality as $\mathbb{R}$? Does every measurable 2-coloring of pairs of reals have a measurable homogeneous set with the same cardinality as $\mathbb{R}$?