Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function satisfying $g'(x) > 0$ for all $x \neq 0$. Prove that $g$ is one-to-one. Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function satisfying $g'(x) > 0$ for all $x \neq 0$.  Prove that $g$ is one-to-one.
Proof:  Case 1: Consider $x_{2} > x_{1}>0$. Then by the Mean Value Theorem, there exists a point $x \in \mathbb{R}$ such that $g'(x) = \frac{g(x_{2}) - g(x_{1})}{x_{2}- x_{1}}$ which means that $g(x_{2}) > g_(x_{1})$ since $g'(x)> 0$ . Thus $g$ is $1-1$ in this case.    
Case 2: Similarily if $x_{1}<x_{2}<0$, then $g(x_{1})<g(x_{2})$. 
Case 3: The only case left to consider is that if $x_{1} \leq 0< x_{2}$
I am not sure how to approach case 3.  Do I use the MVT again or something else.  Also, am I missing any circumstances?

You need the function to be differentiable over the interval $(x_1,x_2)$ and that's why you're considering the case $x_1\leq 0 < x_2$ apart. However, as I've said, the only thing you need to complete the proof is considering continuity of $g$ at $0$. This is a necessary condition (and you have it from differentiability condition). For example, you could have the function 
$$g(x)=\left\{ \begin{align} 2x &\ \ \mbox{   if } x>0\\ x &\ \ \mbox{   if } x<0\\
2 &\ \ \mbox{   if } x=0\end{align}\right.$$
This function satisfy your hypothesis (except for differentiability at $0$) but it's not continuous at $0$, you can see it's not $1-1$ since $g(0)=g(1)$.
