I'm trying to figure out where I'm going wrong with the following proof that $\mathbb{R}$ cannot be well ordered (and thus, ZFC is inconsistent). It's based off a special case of the Erdős–Dushnik–Miller theorem/Ramsey's Theorem for Infinite Graphs as well as the fact the natural linear total order $(<)$ on $\mathbb{R}$ cannot be a well ordering, even for any uncountable subset of $\mathbb{R}$. Ok, here it is:
Assume there exists a well-ordering of $\mathbb{R}$, $(\mathbb{R},\prec)$ and partion the pairs of elements of $\mathbb{R}$ into two sets, $P_1$ and $P_2$ such that:
- $\{a,b\}\in P_1$ iff $a\prec b \iff a<b$, and
- $\{a,b\}\in P_2$ iff $a\prec b \iff b<a$.
$P_1$ and $P_2$ are well defined since $\prec$ and $<$ are both total orders. Pick any $a_0\in \mathbb{R}$. Then since $\mathbb{R}$ is uncountable, at least one of the sets $S_i:=\{x\mid x\in\mathbb{R}\; \text{and } \{a_0,x\}\in P_i\}$ for $i\in\{1,2\}$ is uncountable. Let $S_{i_0}$ be such a set. Pick another number $a_1$ from $S_{i_0}\setminus\{a_0\}$. Then as in the case of $a_0$, there exist uncountable many real numbers $x$ in $S_{i_0}\setminus\{a_0,\;a_1\}$ such that for all of these $x$, all of the pairs $\{a_1,x\}$ belong to one of the $P_i$. Let $S_{i_1}$ be such a set. Then assuming $\mathsf{AC}$ (which we get by working under the assumption that every set can be well ordered), there are uncountably many $a_j$ and associated nested subsets of $\mathbb{R}$, $S_{i_j}$ such that all of the $S_{i_j}$ are uncountable and all the pairs $\{a_j,x\}$ for every $x\in S_{i_j}$ are in $P_{i_j}$ for some $i_j\in\{0,1\}$.
Now there are uncountably many indices $j$ so there must be an uncountable number of the $i_j$ equal to $1$ or $2$. This means that there must be an uncountable number of $a_j$ such for one of the $P_i$, all of the pairs of those $a_j$ are in $P_i$. Let $A$ be the set of these $a_j$. Suppose first that all the pairs of reals in $A$ are in $P_1$. Since we are operating under the assumption that $\prec$ well orders $\mathbb{R}$ and hence $A$, and the fact that $\prec$ agrees with the regular linear order on all pairs of reals in $P_1$ we reach a contradiction: $A$ is uncountable by construction and $<$ does not well order any uncountable subset of $\mathbb{R}$.
Now assume all of the pairs of reals in $A$ are in $P_2$. Define $B:=\{-a\mid\; a\in A,\, a\neq0\}$ where negation here is the usual negation with respect to the linear order. Then by the definition of $P_2$ and the basic property of inequality, for all $b_1,\; b_2\in B$, $b_1<b_2\iff b_1\prec b_2$ and thus in this case the linear order on $B$ also agrees with the supposed well order $\prec$ on $B$ and we get the same contradiction because $B$ is uncountable just as $A$ is.
I can't figure out where I'm going wrong because the proof closely follows the constructive proof for Ramsey's Theorem for infinite graphs. Thanks for your help.