Is it possible to construct an infinite subset of $\Bbb R$ that is not order isomorphic to any proper subset of itself?

There is no such countably infinite set, but there is such a set with size $2^{\aleph_0}$. These are classic results of Dushnik & Miller [Concerning similarity transformations of linearly ordered sets, Bull. Amer. Math. Soc. 46 (1940), 322326]. I also outlined the idea behind this construction in my first MathOverflow answer! 


The answer is yes in ZFC. We can construct a dense infinite set $A\subset\mathbb{R}$ such that the only orderpreserving map $f:A\to A$ is the identity. In particular, $A$ is not orderisomorphic with any proper subset of itself. To see this, note first that any orderpreserving map $f:B\to\mathbb{R}$ defined on a dense set $B\subset\mathbb{R}$ can be extended to a total orderpreserving map $\bar f:\mathbb{R}\to\mathbb{R}$ defined on the closure of $B$, by defining $\bar f(x)=\sup_{y\leq x, y\in B}f(y)$. Further, note that any such monotone map will have at most countably many points of discontinuity, since every discontinuity will be a jump discontinuity. Thus, there are precisely continuum many such orderpreserving functions $\mathbb{R}\to\mathbb{R}$, since any one of them is determined by countably much information about their values on a countable dense set and the information about what their values are on the countably many points of discontinuity. We may therefore enumerate all orderpreserving functions $f_\alpha:\mathbb{R}\to\mathbb{R}$ in a sequence of length continuum, $\alpha\lt\mathfrak{c}$. Let's now build the set $A$ by a transfinite process, making promises at each stage about some reals being definitely in $A$ and other promises about keeping some reals out of $A$, in such a way that we kill off $f_\alpha$ at stage $\alpha$ as a possible orderpreserving map from $A$ to $A$. We may begin at stage $0$ by placing all the rational numbers into $A$, so that it will definitely be dense. Suppose we have carried out our process up to stage $\alpha$, and $f_\alpha$ is the next nonidentity orderpreserving map $\mathbb{R}\to\mathbb{R}$ presented for our consideration. Since $f_\alpha$ is orderpreserving and not the identity, it must be that there is an interval $(a,b)$ with $(f(a),f(b))$ disjoint from $(a,b)$. Since we've made fewer than continuum many promises so far, there must be an $x\in (a,b)$ such that we've made no promises about $x$ or $f_\alpha(x)$. In this case, we place $x$ into $A$ and promise to keep $f_\alpha(x)$ out of $A$. This will prevent $f_\alpha$ from being an orderisomorphism of $A$ to a proper subset of $A$. The end result is that $A$ is dense, but is strongly rigid in the sense that there is no nonidentity orderpreserving map from $A$ to $A$. In particular, $A$ is not orderisomorphic with any proper subset of itself. 


EDIT: The other answers show that my intuition was wrong, and that in fact there is such a linear order in $ZFC$, so this answer (except the bit about determinacy) is superfluous. Although it seems likely that $ZFC$ proves there is no such order, choice will certainly be necesary for such a proof: it is consistent with $ZF$ that $\mathbb{R}$ has infinite, Dedekindfinite subsets, which is exactly what you're asking for. (I'm looking for a reference . . . EDIT: Asaf gives a good reference in a comment, below.) There can be no countable example of such an order, however; such an order can't embed $\mathbb{Q}$, and hence is scattered, and a result (I believe) of Jullien then lets us write it as a finite sum of indecomposable orders; it is then easy to see that the whole order embeds into a proper subset of itself. If we assume the Axiom of Determinacy, then every uncountable set of reals has a perfect subset; this means there is no uncountable example of such an order, and hence by the above fact the answer to your question is no. Under choice, things look a bit more complicated, but I suspect the answer is no; I'll post more if I can figure it out. 

