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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 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 $ϵ>0$ with $B(x,ϵ)\subseteq U$, then choose $c\in C$ with $d(c,x)<ϵ/3$ and rational $r$ satisfyingThe triangle inequality gives $x\in B(c,r)\subseteq B(x,ϵ)\subseteq U$. 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 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 $ϵ>0$,Indeed, $(t,-t)$ in that rectangle requires both $t\ge x$ and $t\le x$. 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 $X$ is separable and $Y$ is not separable, illustrating why the metric assumption in the inheritance statement matters.