Q:

How to show a special continuous bijection in $\mathbb{R}^n$ is a homeomorphism?

Suppose $f: \mathbb{R}^n \to \mathbb{R}^n$ is a continuous bijection such that if $E \subset \mathbb{R}^n$ is bounded, then $f^{-1}(E)$ is bounded.
How can I show that $f^{-1}$ is continuous, and thus $f$ is a homeomorphism?
My reasoning is as follows:

Take $x \in \mathbb{R}^n$, take any compact set $K \subset \mathbb{R}^n$ such that $x \in K$. Then $f^{-1}(K)$ is closed and bounded (since $f$ is continuous and by the assumption), so compact (we are in $\mathbb{R}^n$). Then $f^{-1}(f^{-1}(K)) = f(K)$ is compact, since $f$ is continuous. Thus, since ${f^{-1}}^{-1} = f$ sends closed sets to closed sets, $f^{-1}$ is continuous. 

Is my reasoning right? I feel that there's either a bit of circular reasoning going on, or I'm missing a simpler solution. 

A:

You write $f^{-1}(f^{-1}(K)) = f(K)$, but in general this is not true ! Suppose it is true, then we would get $K=f(K)$ !
Write $g=f^{-1}$
Let $y_0 \in \mathbb R^n$ and let $(y_n)$ be a sequence with $y_n \to y_0$. Then we have to show that $g(y_n) \to g(y_0)$.
The set $E=\{y_1,y_2,y_3,...\}$ is bounded, hence the sequence $(z_n):=(g(y_n))$ is bounded.
It is your turn to show the following: if $z_0$ is an accumulation point of $(z_n)$, then we have $z_0=g(y_0)$.
Consequence: $(g(y_n))$ is bounded and has only one accumulation point, namely $g(y_0)$. Therefore  $g(y_n) \to g(y_0)$.

