I was reading Erdős and Rado's "A Partition Calculus in Set Theory" paper and saw the following proof:
Theorem $\mathbf {23}$: If $n < \omega$ and $ \alpha < \omega \cdot 2$, then $\omega \cdot n \rightarrow (n,\alpha)^2$.
Proof: Let the set $A = \bigcup_{\nu <n} A_{\nu}$ be ordered, $\overline {A_{\nu}} = \omega$ (order type) for $\nu <n$, and $A_{\nu} \subset L(A_{\nu + 1})$ (which means all elements of $A_{\nu}$ are less than all elements of $A_{\nu + 1}$) for $\nu + 1 < n$. Suppose that $[A]^2 = K_0 \cup K_1$; $n \not \in [K_0]$; $\alpha \not \in [K_1]$ (which means they don't have a subset whose underlying set is of the corresponding order type). We want to deduce a contradiction.
By Theorem $1$ (infinite Ramsey theorem), there is, for every $\nu<n$, a set $B_{\nu} \in [A_{\nu}]^{\aleph_0}$ such that $[B_{\nu}]^2 \subset K_{\rho_{\nu}}$, for some $\rho_{\nu} < 2$. Since $n \not \in [K_0]$, we have $\rho_{\nu} = 1$. Let $\lambda < \mu < n$. We define an operator $p_{\lambda \mu}$ as follows. there is at least one set $B \subset B_{\lambda} \cup B_{\mu}$ such that $|B \cap B_{\lambda}| = \aleph_0$; $[B]^2 \subset K_1$. For instance, $B_{\lambda}$ is such a set $B$. Since $\alpha \not \in [K_1]$, we have, for every such $B$, $\omega \leq \overline{B} < \alpha < \omega \cdot 2$. Hence we can choose $B$ such that $\overline{B}$ is maximal. We fix such a $B$ by any suitable convention and put $$P_{\lambda \mu}(B_0,B_1, \ldots, B_{n-1}) = (C_0, C_1, \ldots, C_{n-1}),$$ where $C_{\lambda} = B \cap B_{\lambda}$; $C_{\mu} = B_{\mu} - B$, and $C_{\nu} = B_{\nu}$ for $\nu \neq \lambda, \mu$. Then $\overline{C_\nu} = \omega$; $|C_{\lambda} \cap L_1(x)| < \aleph_0$ for $\nu < n$; $x \in C_{\mu}$ (where $L_1(x) = \{ y < x | \{y,x\} \in K_1 \}$)...
I don't understand why it is the case that $|C_{\lambda} \cap L_1(x)| < \aleph_0$. At first I thought if there were such $x$ then we could add that $x$ to $B$ and contradict the maximality of $\overline{B}$ , but how do we know that the new set is monochromatic? Thank you.