Let $(C, \leq_C)$ be any chain: binary relation $\leq_C$ on $C$ is reflexive, antisymmetric, transitive, and total.
Give $\mathbb{R}^2$ the lexicographic order: for all real numbers $a,b,c,d$, $(a,b) \leq_L (c,d)$ if and only if $a < c$ or ($a = c$ and $b \leq d$).
Question: what are necessary and sufficient conditions for there to be an order-isomorphism from $(C, \leq_C)$ to a subset of $(\mathbb{R}^2, \leq_L)$, i.e., a function $f: C \to \mathbb{R}^2$ with $x \leq_C y$ if and only if $f(x) \leq_L f(y)$?
Motivation: I know what characterizes order-isomorphisms to subsets of $\mathbb{R}$ with its usual order. Birkhoff's Lattice Theory (3rd ed, p. 200), for instance, shows that such an isomorphism exists if and only if $C$ has a countable "order-dense" subset $D$: a countable subset $D$ of $C$ such that for all $a, b \in C \setminus D$ with $a <_C b$ there is a $d \in D$ with $a <_C d <_C b$.
But I find lexicographic orders much more difficult to grasp. My initial guess was that there might be an appealing extension of the order-density-condition to answer my question (for $\mathbb{R}^2$ and perhaps with an inductive argument for $\mathbb{R}^n$), but I have not been able to find one.