If a continuous function is bigger than another at a point, then it is so in an entire neighbourhood...proof? Let $f,g: \mathbb{R}\to \mathbb{R}$, both be continuous at zero. If $f(0)>g(0)$, show that there exists positive $\delta$ such that for all $y,z \in (-\delta , \delta)$, $f(y)>g(z)$.
Attempt at a proof:
I used the obvious choice for $\epsilon >0$, namely $\epsilon:=f(0)-g(0)>0$, and as a $\delta$ I went for the minimum of $\delta_1$ and $\delta_2$ in the definitions of continuity (at zero) of $f$ and $g$ respectively. Then I have both inequalities $|f(x)-f(0)|<f(0) - g(0)$ and $|g(\tilde{x}) - g(0)|<f(0)-g(0)$, hold for $x\in (-\delta , +\delta)$.
Here is where I am stuck, maybe the choice of $\epsilon$ was bad? Any suggestions would be great.

You are almost there. 
You need to create two disjoint neighborhood, centered at $f(o)$ and $g(0)$ and show that each function has its values in the corresponding neighborhood for a small enough interval centered at $0$.
With $\epsilon = (f(0)-g(0))/2$ your goal will be achieved.    
