Let $(X; \preceq)$ be a dense, complete, separable linear order without endpoints.
($\mathbb Q; \le)$ is a countable dense linear order without endpoints and it's a substructure of $(\mathbb R; \le)$.
Step 1. Prove that there is a substructure $(Y; \preceq)$ of $(X; \preceq)$ which is a countable, dense and dense in $(X; \preceq)$ [sic!] linear order without endpoints. Fix an isomorphism
\pi \colon (Y; \preceq) \to (\mathbb Q; \le).
Step 2. Now use that $X$ is complete. Every element of $X$ can be approximated by elements of $Y$ and this approximation translates via $\pi$ to an approximation of some unique element, let's call it $\pi^+(x)$, of $\mathbb R$. Show that
\pi^+ \colon (X; \preceq) \to (\mathbb R; \le)
is an isomorphism. In fact, it's the unique isomorphism $\rho$ between those structures satisfying $\pi \subseteq \rho$.