Showing an open ball is open Let X be a metric space with metric d. 
I'm trying to show that U(xo;e) is an open set. What I note so far is to talk about a subset U that for each xo an element of U, there is a corresponding e > 0 s.t. U(xo;e) is contained in U. 
Does this finish the proof? 
 A: I'm assuming that $U(x_0,e)$ means:
$$U(x_0,e) = \{ x\in X\mid d(x,x_0)\lt e\},$$
that is, the open ball with center in $x_0$ and radius $e$.
I'm also assuming that your definition of open set for a metric space is:

A set $A$ is open if and only if for every $a\in A$ there exists $\varepsilon\gt 0$ such that $U(a,\varepsilon)\subseteq A$.

If this is the case, then what you write does not even begin the proof that for a given $x_0$ and a given $e\gt 0$, $U(x_0,e)$ is open; it only repeats the definition of "open set" in a way that is confusing (using $x_0$ and $e$ for arbitrary points and radii, even though those names are already in use for a specific point and a specific radius).
Instead, what you need to do is the following: to show that $U(x_0,e)$ is open, let $y\in U(x_0,e)$  be a point in the set. What you know is that $d(x_0,y)\lt e$. What you want to show is that there exists a $\varepsilon\gt 0$ such that $U(y,\varepsilon)\subseteq U(x_0,e)$.
What does that mean? A point $z$ is in $U(y,\varepsilon)$ if and only if $d(y,z)\lt \varepsilon$. A point $z$ is in $U(x_0,e)$ if and only if $d(x_0,z)\lt e$. So you want to show that there exists $\varepsilon\gt 0$ with the property that if $d(y,z)\lt \varepsilon$, then $d(x_0,z)\lt e$.
My two word hint for that is: "Triangle Inequality."
