Open neighborhoods in the definition of a manifold At the beginning of Spivak's "A Comprehensive Introduction to Diff. Geom." (p.3),   in the definition of a (topological) manifold $M$, every point $x$ has a neighborhood $U$ that is homeomorphic to $R^n$. Spivak says that $U$ must in fact be an open neighborhood, and that the proof of this is an easy exercise from the Domain Invariance Theorem.
I'm having trouble seeing how to apply the Domain Invariance Theorem here. Once a non-open $U$ is mapped by assumption onto all of $R^n$, everything is trivial about it and the "non-openness" of $U$ in $M$ seems lost. Mapping $R^n$ further by a 1-1 map into R^n will make an open set by Domain Invariance, but I don't see how to reach some kind of contradiction about the original $U$ from that. Would appreciate help - a hint in the right direction even better than a complete answer.
(I tried to pursue directly: suppose some $x$ has a neighborhood $U$ that's not open and homeomorphic to $R^n$. There's an open set $V \subset U$ that contains $x$ and is homeomorphic to $R^n$ (a preimage of an open ball around the image of $x$). So $V$ is homeomorphic to $U$; but I don't know that $V$ is open in $M$, only in $U$. Stuck at this point).
 A: Let $x \in M$ be a point in a manifold, and $U$ a neighborhood of $x$ that's homeomorphic to $R^n$. We will argue that $U$ must be an open set by considering an arbitrary $y \in U$ and proving that $U$ is a neighborhood of $y$, too. 
We know that $y$ has a neighborhood $V$ that's homeomorphic to $R^m$. In Spivak's definition of a manifold, dimension is allowed to vary with the point, so it's possible that $m\ne n$. But we want to ensure that in our specific case $m=n$. We take it as proved that for one point, the dimension cannot vary, and then it follows by a standard connectedness argument that inside a connected subset the dimension cannot vary. $U$ is connected as it is homeomorphic to $R^n$. Additionally, we may take $V$ to be an open set and not just a neighborhood (if not, take an open $V'$ inside $V$ containing $y$, its image in $R^n$ has a ball around the image of $y$, the preimage of the ball is open in $V'$ and therefore in $M$ and homeomorphic to the ball hence to $R^n$).
To sum up, an arbitrary $y$ in $U$ lies in an open $V$ homeomorphic to $R^n$. Let $\phi:U \rightarrow R^n, \psi:V \rightarrow R^n$ be the homeomorphisms. Let $W = U \cap V$. If we prove $W$ open, that establishes an open set around $y$ within $U$ and finishes the proof. We don't yet know that $W$ is open, but at any rate it's open in the subspace topology of $U$, as an intersection of $U$ and open $V$. So $\phi(W)$ is open in $R^n$. Now consider the mapping from $\phi(W)$ to $\psi(W)$ given by composition of $\phi^{-1}$ and $\psi$. It's a continuous 1-1 mapping from an open set in $R^n$ into $R^n$. By the Domain Invariance theorem, $\psi(W)$ is open. By the homeomorphism, $W$ is open as a set in $V$, but since $V$ is itself open in $M$, $W$ is open in $M$ which is what we needed.
