I'm trying to solve a problem in Spivak's A comprehensive introduction to differential geometry. Here, the definition of a manifold is the next
A metric space $X$ is said to be a manifold if every point in $X$ has a neighbourhood homeomorphic to $\mathbb{R}^n$ for some $n$
Given $x\in X$, a neighborhood is just a set $U$ such that $p\in \mathrm{int}U$ (so $U$ is not necessarily open). However, after the enunciation of the Invariance of Domain theorem, he assures that the neighbourhood in the definition of a manifold must in fact be open and that this is an easy consequence of such theorem. This is what I've been trying to prove, although I couldn't succeed doing so.
My attempt so far was the next: Given a point $x$ in the manifold, a neighborhood $U$ of $x$ and a homeomorphism$\phi:U\to \mathbb{R}^n$, I tried to show that $U\subset \mathrm{int} U$, that is, for every $y\in U$ there's an open set $W$ in $X$ such that $y\in W\subset U$. For such a $y$ there exists a neighborhood $V$ of $y$ and a homeomorphism $\psi :V\to\mathbb{R}^m$ for some $m$. Set $A=U\cap \mathrm{int}V$. Since $A$ is open in $U$, $\phi(A)$ is open in $\mathbb{R}^n$, and we have a one-one continuous function $\psi\circ \phi^{-1}: \phi(A)\to \psi(A)$ from an open subset of $\mathbb {R}^n$ to some subset of $\mathbb {R}^m$. Here's where I'm stuck.
I can see that if $\psi(A)$ is open, then we're done. This would follow from ID theorem if $m=n$, but I can't see why this has to be true. From said theorem I was able to show that $n\leq m$ but I couldn't get further.
Could you please help me this? Also, I would appreciate more hints than complete answers. Thank you.