Proof that every open set is a union of neighborhoods. 
Definition: A set $U \in \Bbb R^n$ is open if every point $x \in U$ has a neighborhood
  contained in the set....
  Question: Show that every open set $U \in \Bbb R^n$ is a union of neighborhoods of points of $U$.

This sounds like a tautology to me.  What is there for me to prove?
 A: This is a standard construction.  (You will use it a lot.)
Let $U$ be an open set.  For each $x \in U$, let $N_x$ be the open neighborhood in $U$ containing $x$, which is guaranteed to exist by the definition.  Consider $\mathcal{N} = \bigcup_{x \in U} N_x$.  Since every $x \in N_x \subset \mathcal{N}$, we find $U \subset \mathcal{N}$.  Since every $N_x \subset U$, we find $\mathcal{N} \subset U$.  Therefore, $U = \mathcal{N}$ is a union of neighborhoods of points of $U$.
A: Let $U\subseteq\Bbb R^n$ be open. By definition, for every $x\in U$, there exists a neighborhood $U_x\subseteq U$ such that $x\in U_x$. It follows that
$$
U=\bigcup_{x\in U}U_x
$$
as required.
A: I think the definition and the question are the same thing here.
If you ignore the definition and consider the question then here is a method to solve it.
A set is open if and only if it is equal to its interior i.e every point of the set is an interior point.
Now a point "x" is an interior point of a set A if there exists an open set containing the point "x" which is contained in A.
Now coming back to your question: Let U be an open set. Then for every "x" in U there exists an open set of "x" contained in U. Take the union of all such open sets. Then it is easy to prove that this union is equal to the set U. Basically show that U is a subset of this union and vice-versa.
