Is it possible to construct an infinite subset of $\Bbb R$ that is not order isomorphic to any proper subset of itself? Is it possible to construct an infinite subset of $\Bbb R$ that is not order isomorphic to any proper subset of itself?
 A: 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, Dedekind-finite 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.
A: The answer is yes in ZFC. We can construct a dense infinite set
$A\subset\mathbb{R}$ such that the only order-preserving map
$f:A\to A$ is the identity. In particular, $A$ is not
order-isomorphic with any proper subset of itself.
To see this, note first that any order-preserving map
$f:B\to\mathbb{R}$ defined on a dense set $B\subset\mathbb{R}$ can
be extended to a total order-preserving 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
order-preserving 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 order-preserving 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
order-preserving 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 non-identity
order-preserving map $\mathbb{R}\to\mathbb{R}$ presented for our
consideration. Since $f_\alpha$ is order-preserving 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
order-isomorphism 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 non-identity order-preserving map from $A$
to $A$. In particular, $A$ is not order-isomorphic with any proper
subset of itself.
A: 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), 322-326]. I also outlined the idea behind this construction in my first MathOverflow answer!
