The first 'easy exercise' in Spivak's differential geometry book I am a 'lowly physicist' (actually, only a prospective physicist) with little to no background in formal mathematics. Recently I decided it's time to get serious, so I started reading Spivak's 'Comprehensive Introduction to Differential Geometry'. The first chapter, on manifolds, starts off with the definition of a manifold:
Definition: A manifold is a metric space $M$ with the property that, for any point $x\in M$, there exists a neighborhood $U$ of $x$ and some integer $n\geq 0$ such that $U$ is homeomorphic to $\mathbb{R}^n$.
On the third page, Spivak argues that $U$ must be an open set. He has already demonstrated that we can always choose $U$ to be open. He gives, without proof, the 'Invariance of domain' theorem:
Theorem: If $U\subset \mathbb{R}^n$ is open and $f: U\to \mathbb{R}^n$ is one-one and continuous, then $f(U)\subset \mathbb{R}^n$ is open. (it follows that $f(V)$ is open for any open $V\subset U$, so $f^{-1}$ is continuous and $f:U\to f(U)$ is a homeomorphism.)
Now, Spivak states that it is 'an easy exercise' to show that this theorem implies that $U$ from the above definition must be open. 
I have tried to look up and understand all the relevant definitions. What I (think I) have understood so far is the following. From the property that $U$ is homeomorphic to $\mathbb{R}^n$ it follows immediately that there is also a homeomorphism (i.e. a continuous and one-one (?) function) from $\mathbb{R}^n$ to $U$. If we denote the homeomorphism from $U$ to $\mathbb{R}^n$ by $f$, it is enough to show that $f(U)$ is open to conclude that $f$ is open by the invariance of domain theorem. However, it is not clear to me that this must be the case. Am I missing a subtlety in a definition? Or is there some nontrivial step here? 
 A: The question is to prove that if $U \subset M$ is homeomorphic to $\mathbb{R}^n$ then $U$ is open. Choose a homeomorphism $f :U \to \mathbb{R}^n$ (yes, homeomorphisms are one-to-one, onto, and they are continuous with continuous inverse; also, to be explicit, the topology on $U$ is the subspace topology, whose open sets are the intersections of $U$ with open subsets of $M$).
Take a point $x \in U$. We can choose an open subset $V_x \subset M$ containing $x$ and a homeomorphism $f_x : V_x \to \mathbb{R}^n$. The set $V_x \cap U$ is open in the subspace topology on $U$. Therefore $f(V_x \cap U) \subset \mathbb{R}^n$ is open, and it contains $f(x)$. There is an open ball $B \subset \mathbb{R}^n$ centered on $f(x)$ such that $B \subset f(V_x \cap U)$. The function $f_x \circ f^{-1}$ restricted to $B$ is one-to-one and continuous, and so by applying the Invariance of Domain Theorem its image $f_x \circ f^{-1}(B)$ is an open subset of $f_x(V_x)=\mathbb{R}^n$. It follows that $f^{-1}(B) = f_x^{-1} \circ f_x \circ f^{-1}(B)$ is an open subset of $V_x$ which is an open subset of $M$. Therefore $f^{-1}(B)$ is an open subset of $M$. Also, by construction $x \in f^{-1}(B) \subset U$. Since $x \in U$ is arbitrary, $U$ is open.
