Prove that if all set homeomorphic to X is closed, then X is compact. Let $X⊂\mathbb{R}^n$. Prove that if all set (subsets of $\mathbb{R}^n$) homeomorphic to $X$ is closed, then $X$ is compact.
I need some help to think on this problem. There is a similar question, but the first implication is "...homeomorphic to $X$ is bounded...". For this variation of my question, the solution uses a function like
$$f(x) = \frac{x-x_0}{|x-x_0|^2}$$
I have a felling that I need a funtion and try to build some contradiction...but I don't know how. Is this function above useful for my problem?
Thank you for any answer.
 A: As @Rob Arthan suggested, I am writing my comments as an answer to remove the question from the unanswered queue. Steps are below:
$(1)$ Let $\Bbb B^n:=\{x\in \Bbb R^n:|x|<1\}$. Consider the map $f\colon \Bbb B^n\to \Bbb R^n$ defined by $f(x):=\frac{x}{1-|x|}$. Note that $f$ is a homeomorphism with inverse $f^{-1}\colon \Bbb R^n\ni z\longmapsto \frac{z}{1+|z|}\in \Bbb B^n$.
$(2)$ Now, let $Y:=f^{-1}(X)$. From the assumption, $X$ is closed in $\Bbb R^n$. Since $f^{-1}$ is a homeomorphism, $Y$ is closed in $\Bbb B^n$.
$(3)$ Let $A$ be a non-compact closed subset of $\Bbb B^n$. Actually, $\overline A\subseteq \{x\in \Bbb R^n:|x|\leq1\}$. Here, the closure $\overline A$ is taken in $\Bbb R^n$. So, $\overline A$ is a compact subset of $\Bbb R^n$ as it is closed in $\Bbb R^n$ as well as a bounded subset of $\Bbb R^n$. Note that $\overline A\cap \Bbb B^n =A$ as $A$ is closed in $\Bbb B^n$. Also, note that $\overline A\backslash \Bbb B^n\neq \varnothing$, otherwise $A$ would be compact subset $\Bbb B^n$. Therefore, there is a sequence of points of $A$ converging to some point $\overline A\cap \partial \Bbb B^n$.
$(4)$ Now, if $X$ were non-compact, then $Y(\cong X)$ would be a non-compact closed subset of $\Bbb B^n$, and then $Y$ would have a limit point in $\partial \Bbb B^n$ by $(3)$, i.e.,   $Y$ wouldn't be a closed subset of $\Bbb R^n$, a contradiction to the hypothesis that every homeomorphic copy(in $\Bbb R^n$) of $X$ is closed in $\Bbb R^n$.
