f is continuous at a if and only if osc(f,a)=0 So I'm trying to prove this:
Let f:A-->R be bounded. Then f is continuous at a if and only if osc(f,a)=0. 
I have the proof assuming f is continuous and showing osc(f,a)=0. However, I can't prove the converse, though I know it is similar.
Here's the first part:
Let f be continuous at a. For every number ϵ>0 we can choose a number δ>0 so that |f(x)-f(a)|<ϵ for all x∈A with |x-a|<δ. Thus M(a,f,δ)-m(a,f,δ)≤2ϵ, Since this is true for every ϵ,we have osc(f,a)=0.

To prove the converse of what you showed, suppose $osc(f,a)=0$.  Then given $\epsilon>0$, choose $\delta_0$ such that $\delta<\delta_0 \implies M(a,f,\delta)-m(a,f,\delta) < \epsilon$.  Then $|x-a|<\delta_0 \implies |f(a)-f(x)| < M(a,f,\delta)-m(a,f,\delta) < \epsilon$.
