Is Invariance Of Domain (Fundamentally) Equivalent To Every Continuous Bijection $\mathbf{R}^n\to\mathbf{R}^n$ Is Open? I found this comment

I would also like to point out that invariance of domain and the statement that every continuous bijection from $\mathbf{R}^n$ to $\mathbf{R}^n$ is a homeomorphism are equivalent. Therefore I would not expect there to be a proof that avoids this machinery.

but couldn't prove it myself using fundamental methods, i.e. no algebraic topology.  As far as I can tell, invariance of domain is equivalent to the statement restricted to open balls in $\mathbf{R}^n$.  But from there, I don't see how to prove this using only the statement about continuous bijections and no algebraic topology.  Clearly, the preimage of an open ball is open, but how can I conclude that it's homeomorphic to an open ball?
 A: Let's call a topological space $X$ domain invariant, if it fullfills the usual condition of the theorem of domain invariance, i.e. if $U, V$ are homeomorphis subsets of $X$, and $U$ is open, then also $V$ is open.
$X$ fullfills (*), if every continuous bijection $f: X \rightarrow X$ is a homeomorphism (sorry, no better notation came up to my mind).
By a standard argument we have:
If X is doamin invariant, T2 and locally compact, then $X$ fullfills (*).
Proof. Let $f: X \rightarrow X$ be a continuous bijection. Of course, it suffices to show that $f$ is open: Let $U$ be open in $X$, $x \in U$. By local compactness pick an open $V$ such that $x \in V \subset \overline{V} \subset U$ and $\overline{V}$ compact. Then $f|\overline{V}: \overline{V} \rightarrow f(\overline{V})$ is continuous, bijective, hence a homeomorphism. Thus, $f|V: V \rightarrow f(V)$ is a homeomorhism, and $ V \cong f(V)$. Since $X$ is domain invariant, $f(V)$ is open. Therefore $f(U)$ is open.
This might be interpreted as "$\Rightarrow$" of the above statement (if $X$ is T2, locally compact).
However, "$\Leftarrow$" is not true in general:
Of course, if $X$ is compact, T2 then $X$ fullfills (*), but need not be domain invariant, as the unit interval shows.
