Equivalence of Definitions of Generalized Ordered Space/Definition 2 implies Definition 1

Theorem
Let $\left({X,\preceq}\right)$ be a totally ordered set.

Let $\tau$ be a topology over $X$.

Let $\left({X',\preceq',\tau'}\right)$ be a linearly ordered space.

Let $\phi:X \to X'$ be an order embedding and a topological embedding.

Then $\tau$ has a basis consisting of convex sets.

Proof
Let $x \in U \in \tau$.

Then by the definition of topological embedding, $\phi(U)$ is an open neighborhood of $\phi \left({x}\right)$ in $\phi \left({X}\right)$ with the subspace topology.

Thus by Basis for Topological Subspace and the definition of the order topology, there is an open interval or open ray $I' \in \tau'$ such that
 * $\phi \left({x}\right) \in I' \cap \phi \left({X}\right) \subseteq \phi \left({U}\right)$.

Since $I'$ is an interval or ray, it is convex in $X'$ by Interval is Convex or Ray is Convex, respectively.

Then $x \in \phi^{-1} \left({I'}\right) = \phi^{-1}\left({I' \cap \phi \left({X}\right)}\right) \subseteq \phi^{-1}\left({\phi \left({U}\right)}\right)$.

Since $\phi$ is a topological embedding, it is injective, so:
 * $x \in \phi^{-1} \left({I'}\right) \subseteq U$.

By Inverse Image of Convex Set under Monotone Mapping is Convex:
 * $\phi^{-1}\left({I'}\right)$ is convex.

Thus $\tau$ has a basis consisting of convex sets.