Is there an infinite subset of $\Bbb{R}$ that is not homeomorphic to any of its proper subsets? Clearly, any finite subset of $\Bbb{R}$ is not homeomorphic to any of its proper subsets by mere cardinality. Also, there are infinite topological spaces that also share this property, e.g. $S^n$. So the infinitude condition and the subspace of $\Bbb R$ condition are crucial.

**What I know.**

To make it more concise, let's say that $(X,\tau)$ is topologically Dedekind-infinite if there is some proper subset $S\subsetneq X$ such that, equipped with the restricted topology, is homeomorphic to $X$. The assertion "$X$ is topologically Dedekind-infinite" will be denoted as ${\sf TInf}(X)$ and the assertion "$X$ is (Dedekind-)infinite as a set" will be denoted as ${\sf Inf}(X)$.

If it is true that $\exists X\subseteq \Bbb R:\lnot{\sf TInf}(X)\land{\sf Inf}(X)$, then $\exists X\subseteq\Bbb R:\lnot{\sf TInf}(X)$ satisfying the following conditions:

- $X$ has empty interior (hence, is totally disconnected)
- $X$ is bounded
- $X$ has no isolated points
- $X$ isn't closed
- $X$ isn't countable

The first condition follows from the fact that, if $X$ has an interior point, it must contain an open interval $I\subsetneq X$. So, if we consider any homeomorphism $f:\Bbb{R}\to I$, we have that $f\vert_{X}:X\to I\subsetneq X$ is an embedding that would imply ${\sf TInf}(X)$. The totally disconnected part follows from the fact that, for any $x_1<x_2$ in $X$, there has to exist some $x_1<z<x_2$ not in $X$ since $X$ can't contain open intervals.

The second condition follows from the fact that, if $\lnot{\sf TInf}(X)$ and $\Phi:\Bbb R\to(0,1)$ is a homeomorphism, then $\lnot{\sf TInf}(\Phi[X])$ and $\Phi[X]$ is bounded.

The third condition follows from two facts. The first one is that ${\sf Inf}(X')\Rightarrow{\sf TInf}(X)$ where $X'$ is the derived set of $X$ (the set of isolated points of $X$). This is because, if $\{x_n\}_{n\in\Bbb N}\subseteq X$ is a set of distinct isolated points, then $f:X\to X$ defined as $x_n\mapsto x_{n+1}$ (and $x\mapsto x$ elsewhere) is a non-surjective embedding $X\to X$ implying ${\sf TInf}(X)$. The other fact is that, if $x_0\in X$ is an isolated point, then ${\sf TInf}(X\setminus\{x_0\})\Rightarrow{\sf TInf}(X)$ since any non-surjective embedding $f:X\setminus\{x_0\}\to X\setminus\{x_0\}$ can be extended as a non-surjective embedding $f^\sharp:X\to X$ with $x_0\mapsto x_0$ and $f^\sharp\vert_{X\setminus\{x_0\}}=f$. By contraposition, the $\lnot {\sf TInf}$ condition is preserved when removing finitely many isolated and, since there will always be finitely many by the previous fact, it will be preserved when removing all its isolated points.

The fourth condition follows from the fact that, if $X$ satisfies the first three properties being nonempty, being closed would also imply that $X$ is compact. Thus, by the characterization of the Cantor space, $X$ would have to be homeomorphic to the Cantor space which is ${\sf TInf}$ (it is self-similar).

The fifth condition follows from the fact that, if $X$ was countable, then $X$ would be a countable metrizable space with no isolated points so, by Sierpiński's topological characterization of $\Bbb Q$, it would be homeomorphic to $\Bbb Q$ which is ${\sf TInf}$.

**What I tried.**

I basically tried constructing such an infinite set following the philosophy of the answers given in this MO question to no avail since I don't know if it is possible to encapsulate the desired condition as a family of $\mathfrak{c}$ simple conditions (as Joel David Hamkins did).