This is an update to an [older question](https://mathoverflow.net/questions/323629/does-every-connected-t-2-space-x-have-a-proper-subspace-homeomorphic-to-x).


Is there a contractible $T\_2$-space $(X,\tau)$ on more than $1$ point such that no proper subspace of $X$ is homeomorphic to $X$?



Contractible $T_2$-space without subspace homeomorphic to it
An example of such contractible (compact metrizable) space can be constructed as follows.


Let $K$ be [the Cook continuum](http://matwbn.icm.edu.pl/ksiazki/fm/fm60/fm60123.pdf). It has the property that any continuous map $f:C\to K$ defined on a subcontinuum $C\subset K$ is either constant or the identity inclusion. Take any countable family $(K\_{n,m})\_{n,m\in\omega}$ of pairwise disjoint non-degenerate subcontinua of $K$. For every $n,m\in\omega$ let $\hat K\_{n,m}=K\_{n,m}\times [0,1]/(K\_{n,m}\times\{0\})$ be the cone over $K\_{n,m}$ with vertex $v\_{n,m}$. 


Let $X\_0=\hat K\_{0,0}$. For every $n\in\mathbb N$ we shall construct a compact metrizable contractible space $X\_n$ that contain the space $X\_{n-1}$ as a deformation retract. Assume that for some $n\in\mathbb N$ a space $X\_{n-1}$ has been contructed. Fix any function $f\_n:\omega\to X\_{n-1}$ with dense image $f\_n(\omega)$ in $X\_{n-1}$. Let $X\_n$ be the quotient space of the disjoint union $X\_{n-1}\cup\bigcup\_{m\in\omega}\hat K\_{n,m}$ by the equivalence relation that identifies the vertex $v\_{n,m}$ of each cone $\hat K\_{n,m}$ with the point $f\_n(m)$. The space $X\_n$ has a natural compact metrizable topology such that $X\_{n-1}$ is a reformation retract of $X\_n$. This compeletes the inductive step.


Then the inverse limit $X=\varprojlim X\_n$ is a contractible compact Hausdorff space such that every proper subspace $P$ of $X$ is not homeomorphic to $X$. This is because $P$ is either non-compact or not dense in $X$ and hence contains no subset homeomorphic to the continuum $K\_{n,m}$ for suitable (large) numbers $n,m$.


