Refuted: separability is hereditary

The half-open intervals cover $\mathbb{R}$, and if two contain $x$, then $[x,c)$ lies in their intersection for some $c>x$; hence [F1] makes them a basis. The rational grid $\mathbb{Q}\times \mathbb{Q}$ is at most countable by [L1] and [L2], and density of $\mathbb{Q}$ makes it meet every nonempty basic lower-limit rectangle, so it is dense in $P$.

For each $x\in \mathbb{R}$, the basic rectangle $[x,x+1)\times [-x,-x+1)$ meets $A$ only in $(x,-x)$; hence $A$ is discrete in its subspace topology.

The map $x\mapsto (x,-x)$ is a bijection from the uncountable set $\mathbb{R}$ onto $A$, so a dense subset of the discrete space $A$ must be all of $A$ and cannot be at most countable.

Thus $P$ is separable by step 1.1 but has the nonseparable subspace $A$ by step 2.1, refuting heredity of separability.