Isomorphic subsets of countable total orders Suppose $(\Omega,\leq)$ is a totally ordered set, with $\Omega$ infinite and countable. If $S$ is an infinite subset of $\Omega$, then $(S,\leq)$ denotes the induced totally ordered set.
Are there examples of such $(\Omega,\leq)$ for which no $(S,\leq)$ is order isomorphic to $(S',\leq)$ with $S' \subset S \ne S'$ ?
 A: No, no such $(\Omega,\leq)$ exists. Let $(\Omega,\leq)$ be any countably infinite linear order. Then there is a subset $S\subseteq \Omega$ such that $(S,\leq)$ is isomorphic to either $\mathbb{N}$ or $\mathbb{N}^*$, where the latter is $\mathbb{N}$ with the usual order reversed (see below for a proof). In the first case, let $S'$ be $S$ without the least element, and in the second case, let $S'$ be $S$ without the gretest element. In either case, we have $S'\subsetneq S$, but $S'\cong S$. 
Why does such a set $S$ exist? I think the easiest way to prove it is by Ramsey's theorem. Let $(a_n)_{n\in \mathbb{N}}$ be an enumeration of $\Omega$. We color a pair $(a_n,a_m)$ with $n<m$ red if $a_n\leq a_m$ in $\Omega$ and blue if $a_m \leq a_n$ in $\Omega$. By Ramsey's theorem, there is an infinite homogeneous subset $H\subseteq \Omega$. If all pairs are colored red, then the map $a_n\mapsto n$ embeds $H$ in $\mathbb{N}$, and every infinite subset of $\mathbb{N}$ is order-isomorphic to $\mathbb{N}$. Similarly, if all the pairs are colored blue, then $H$ embeds in $\mathbb{N}^*$. 

I believe your original question, before the edit, was: 

Is there a countably infinite linear order $(\Omega,\leq)$ which is not isomorphic to any of its proper suborders? 

This is a much more interesting question, and the answer can be found here: https://mathoverflow.net/questions/131933/is-it-possible-to-construct-an-infinite-subset-of-bbb-r-that-is-not-order-iso 
The paper of Dushnik and Miller linked in the answer by François G. Dorais proves: 


*

*Every countably infinite linear order is isomorphic to a proper suborder.

*There is a suborder $\Omega\subseteq \mathbb{R}$ of cardinality $2^{\aleph_0}$ such that $\Omega$ is not isomorphic to any proper suborder. 
