Is there an elementary proof for the fact that homeomorphism preserves open sets in Euclidean Spaces? Let $f:A \subset \mathbb{R}^n \to \mathbb{R}^n$ be a homeomorphism onto its image, where $A$ is an arbitrary subset of $\mathbb{R}^n.$ I want to show that for every open set $U \subset A$ (where $U$ is open in $\mathbb{R}^n$) the set $f(U)$ is open.
I saw related questions where other users mention about the Invariance of domain theorem. I know that such theorem has a hard proof. But note that the hypotesis of that theorem is "$f$ is a continuous bijection" and I am asumming something a little stronger: "$f$ is an homeomorphism".
I'm trying to prove this theorem by using the following proposition:

Proposition 1. $f:A \subset \mathbb{R}^n \to \mathbb{R}^m$ is continuous if and only if for every open set $U \subset \mathbb{R}^m$, there exist an open subset $V\subset \mathbb{R}^n$ such that $f^{-1}(U)=V\cap A.$

It is possible doing it using the proposition 1?
This is my attempt:
Since $f$ is a homeomorphism, then $f^{-1}:f(A) \to \mathbb{R}^n$ is continuous. If $U\subset \mathbb{R}^n$ is open, then by Proposition 1 we have some open set $V \subset \mathbb{R}^n$ such that $(f^{-1})^{-1}(U)=f(U)=V\cap f(A). $ then... I was trying to proof that $V\cap f(A)$ is open using the continuity of $f,$ but I get stuck in this step.
Can someone please help me with this? Thanks in advance.
 A: A homeomorphism is a bicontinuous bijection (onto its image -- redundant, since that is always what "$f:X \rightarrow Y$ is a [particular map class]" means).  As it is a bijection, it has an inverse.  Bicontinuous means that both the map and its inverse are continuous; equivalently, the map is both an open map and a continuous map.  (... and both properties hold for the inverse map.)
Proposition 1 is too general to prove the theorem.  Proposition 1 must write "$V \cap A$" since all that can be promised is that the preimage of an open set is relatively open in $A$.  As a very easy example, take $A$ closed in $\Bbb{R}^n$ and for $U$ take any open set containing $f(A)$.  Then $f^{-1}(U) = A$, so is closed in $\Bbb{R}^n$ and relatively open in $A$.
A concrete example when $n < m$:  Let $A = [0,1]$, a closed set in $\Bbb{R}^1$, and let $f:A \subset \Bbb{R}^1 \rightarrow \Bbb{R}^2 : (x) \mapsto (x,0)$.  Then take for $U$ the open ball centered at $(0,0)$ of radius $2$.  Since $f(A) \subset U$, we can take any open set in $\Bbb{R}$ that contains $A$ as $V$, for instance, $V = (-1,3)$.  Then $f^{-1}(U) = V \cap A = A$, but that intersection is not open (in $\Bbb{R}$); that intersection is relatively open in $A$.
A concrete example when $n > m$: Let $A = [0,1] \times [0,1]$, a closed set in $\Bbb{R}^2$, and let $f:A \subset \Bbb{R}^2 \rightarrow  \Bbb{R}^1: (x,y) \mapsto (x)$.  Then take for $U$ the open ball centered at $(0)$ of radius $2$.  Since $f(A) \subset U$, we can take any open set in $\Bbb{R}^2$ that contains $A$ for $V$, for instance $(-1,3) \times (-1,3)$.  Then $f^{-1}(U) = V \cap A = A$, but that intersection is not open in $\Bbb{R}^2$; that intersection is relatively open in $A$.
From these two examples, we see that proposition 1 must have "$V \cap A$" in its conclusion when $n \neq m$.
Something special happens when $n = m$ and that something special is not captured by proposition 1.  In particular, the homeomorphism in the proposition for $n \neq m$ "crushes" subsets of the larger space to subsets of the smaller space (along either $f$ or $f^{-1}$ depending, respectively, on whether $n$ or $m$ is larger).  (Think about this in the context of invariance of domain: the image of a homeomorphism looks like a possibly folded, twisted, and distorted embedding of the domain.  When $n<m$, the image cannot be open because the image cannot contain an open ball in $\Bbb{R}^m$.  When $n > m$, the same argument applies to the inverse.)  When $n = m$, there are no "directions" along which a homeomorphism is permitted such crushing, but proposition 1 does not have a separate conclusion for this case, so does not capture this additional constraint when $n = m$.
Of course, anything provable by proof system $P$ can be proven in $P \cup \text{Prop. 1}$, by ignoring proposition 1.  This can't really be said to meet your criterion "possible doing it using the proposition 1".
A: There is no need to carry $f(A)$ around since you know it is $\mathbb{R}^n$.
$f\colon A\to\mathbb{R}^n$ is a homeomorphism, so $f^{-1}\colon\mathbb{R}^n\to A$ is also a homeomorphism.  In particular, $f^{-1}$ is continuous and so $f(U)=(f^{-1})^{-1}(U)$ is open.
A: I would say this more or less follows from the definition of homeomorphism, or at least that's how one should think about it after they've seen a proof at least once. But in general, homeomorphism means topological isomorphism, so "of course" $f$ will send open sets to open sets. I'd argue that a "proof" of this fact makes the claim a little less believable, but never mind, we can prove it.
To "prove" this, it helps to look at things a little more generally. Let $B$ be the image of $f$. Now forget where $A$ and $B$ came from: they are both topological spaces and $f\colon A\to B$ is a homeomorphism. The inverse $g=f^{-1}$ is continuous, so if $U\subset A$ is open, then $g^{-1}(U) = (f^{-1})^{-1}(U) = f(U)\subset B$ is open, by the definition of continuity.
A: Brouwer theorem for invariance of domains says that if n is a natural number and U is open in $R^n$ and $f:U->R^n$ is a continuous injection, then f is an open map.
