= Solution
A <basis of a topology> is a family $\mathcal B$ of open sets such that every open set is a union of members of $\mathcal B$. Equivalently, whenever $x\in U$ and $U$ is open, there is $B\in\mathcal B$ with $x\in B\subseteq U$. For a <metric space>, all open balls form a basis. For the <product topology> on $X\times Y$, the rectangles $U\times V$ with $U,V$ open in their respective factors form a basis; rectangles made from factor bases also suffice.
If $C\subseteq X$ and $D\subseteq Y$ are countable <dense subsets>, then $C\times D$ is countable. Every nonempty basic rectangle meets it, since $U\cap C$ and $V\cap D$ are nonempty. Every nonempty open set contains such a rectangle, proving that \b[a product of two separable spaces is separable].
Suppose a <metric space> has a countable dense set $C$. The balls $B(c,r)$ with $c\in C$ and positive rational $r$ form a countable basis. Indeed, given $x\in U$ open, choose $\epsilon>0$ with $B(x,\epsilon)\subseteq U$, then choose $c\in C$ with $d(c,x)<\epsilon/3$ and rational $r$ satisfying
$$
d(c,x)<r<\epsilon-d(c,x).
$$
The triangle inequality gives $x\in B(c,r)\subseteq B(x,\epsilon)\subseteq U$. Conversely, from any countable basis select one point of each nonempty member. These points form a countable <dense subset>, because every nonempty open set contains a basis member. This direction holds for arbitrary topological spaces. Thus a <metric space> is separable exactly when it is a <second-countable space>. Intersecting a countable basis with a subspace gives a countable basis for its <subspace topology>; selecting points as above shows that \b[every subspace of a separable metric space is separable].
For the <Sorgenfrey line>, $\mathbb Q$ meets every nonempty interval $[a,b)$, so it is a countable <dense subset>. In the <Sorgenfrey plane>, however, for any $x\in\mathbb R$ and $\epsilon>0$,
$$
\bigl([x,x+\epsilon)\times[-x,-x+\epsilon)\bigr)\cap Y=\{(x,-x)\}.
$$
Indeed, $(t,-t)$ in that rectangle requires both $t\geq x$ and $t\leq x$. Every singleton is therefore open in the <subspace topology> on $Y$, which is the <discrete topology>. A dense subset of a discrete space must contain every point. Since $Y$ is in bijection with the uncountable set $\mathbb R$, it has no countable dense subset. Hence \b[$X$ is separable and $Y$ is not separable], illustrating why the metric assumption in the inheritance statement matters.
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