The antidiagonal $\{(x,-x)\}$ is an uncountable discrete subspace of the Sorgenfrey plane, so having a countable dense subset is not a hereditary property

Refuted: that the property "has a countable dense subset" is hereditary (Hereditary, open-hereditary and closed-hereditary properties of topological spaces, Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

Witness. In the Sorgenfrey plane $S \times S$ (The Sorgenfrey plane: the product of two half-open-interval lines has the rectangles $[a,b) \times [c,d)$ as a basis and $\mathbb{Q} \times \mathbb{Q}$ as a countable dense subset), which has the countable dense subset $\mathbb{Q}\times\mathbb{Q}$, take the antidiagonal

$L \;:=\; \{\, (x,-x) : x \in \mathbb{R} \,\}$

with the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then:

1. $L$ is discrete: for every $x$, the basic rectangle $[x,\ x+1) \times [-x,\ -x+1)$ meets $L$ exactly in $\{(x,-x)\}$, so every singleton of $L$ is open in $L$ and the subspace topology is the discrete one (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).
2. $L$ is uncountable (Finite, countably infinite, countable, uncountable), being in bijection with $\mathbb{R}$ ($\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)).
3. The only dense subset of $L$ is $L$ itself, since in a discrete space every subset is closed. So $L$ has no countable dense subset, although the space it sits inside has one.

The word separable is not used: it is not defined at this point in the reading order, and the three claims above say in full what it would abbreviate.

• The rectangle in step 1.1 is chosen with both corners at the point. Any basic rectangle $[x,b) \times [-x,d)$ with $b > x$ and $d > -x$ would do, and the computation is the same: the first factor forces $t \ge x$ and the second forces $t \le x$. It is the half-openness on the left in both coordinates, together with the reversal of the sign in the second, that isolates the point.

• The property is open-hereditary, and that is the sharp statement. By claim 4 of For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is $\overline{A}^{X} \cap S$, while the interior only contains $\operatorname{int}^{X}(A) \cap S$, with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ a dense subset of a space traces to a dense subset of every open subspace, so "has a countable dense subset" passes to open subspaces. The antidiagonal is not open in $S \times S$, and the failure above shows that the hypothesis in that claim cannot be dropped.

• Nothing here needs a choice principle. The surjection $L \to \mathbb{R}$ is written down, the rectangles are written down, and the only nonconstructive ingredient is $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874), whose own proof is choice free.