Assuming choice, normality is not productive: the normal lower-limit line has a nonnormal square

Given: The Axiom of Choice and the lower-limit line $L$.

The Axiom of Choice supplies a choice function for every family of nonempty sets, hence for every countably indexed family, which is the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)).

Under the Axiom of Countable Choice, the lower-limit line is normal (Assuming countable choice, the lower-limit line is normal).

Jones's lemma injects $\mathcal{P}(D)$ into $\mathcal{P}(E)$ when a normal space has closed discrete $D$ and dense $E$ (Jones's bound: under choice, a closed discrete subspace of a normal space cannot have more subsets than a dense set has subsets).

The plane $L^2$ has $E={\mathbb{Q}}^2$ at most countable and $D=\{(x,-x):x\in \mathbb{R}\}$ closed discrete with $D\approx \mathbb{R}$ (The lower-limit plane has a countable dense set and a closed discrete antidiagonal of size $\mathrm{\mid }\mathbb{R}\mathrm{\mid }$).

Cantor's theorem gives no injection $\mathcal{P}(\mathcal{P}(\mathbb{N}))\to \mathcal{P}(\mathbb{N})$, and Schröder-Bernstein turns injections both ways into a bijection (Assuming the Axiom of Choice, $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$, and Cantor's theorem in cardinal form: $\kappa <2^{\kappa }$, The Schröder-Bernstein theorem).

The ternary Cantor-set coding injects $\mathcal{P}(\mathbb{N})$ into $\mathbb{R}$, while $x\mapsto \{q\in \mathbb{Q}:q<x\}$ injects $\mathbb{R}$ into $\mathcal{P}(\mathbb{Q})$; a rational between distinct reals makes the latter map injective, and $\mathbb{Q}\approx \mathbb{N}$ (The Cantor set is exactly the set of ${\sum }_{k\ge 1}{a}_k{3}^{-k}$ with every $a_k\in \{0,2\}$, and this gives a bijection with $\{0,1{\}}^{\mathbb{N}}$, ℚ is dense in every Archimedean ordered field, $\mathbb{Q}$ is countably infinite).