I'm stuck on how to do this proof.

I thought I might start like this...Originally Posted byGaugan. Introduction To Analysis, 5th ed, p105

Choose $\displaystyle \epsilon>0$. By definition of continuity, for each $\displaystyle z_{\alpha}\in R$, there exists $\displaystyle \delta_{\alpha}>0$ such that if $\displaystyle |z-z_{\alpha}|<\delta_{\alpha}$ and $\displaystyle z\in R$, then $\displaystyle |f(z)-f(z_{\alpha})|<\epsilon$.

That way, if I can somehow show that $\displaystyle \inf\{\delta_{\alpha}\}>0$, then I could just finish thusly:

Choose $\displaystyle \delta=\inf\{\delta_{\alpha}\}$. Then for each $\displaystyle x\in R$ and $\displaystyle y\in\{z_{\alpha}\}=R$, if $\displaystyle |x-y|<\delta$ and $\displaystyle x\in R$, then $\displaystyle |f(x)-f(y)|<\epsilon$.

But how do I show that $\displaystyle \inf\{\delta_{\alpha}\}>0$? Or, if that's impossible, then perhaps instead of setting $\displaystyle \delta_{\alpha}>0$, I could set it to $\displaystyle \delta_{\alpha}\in(0,h/4)$, and then somehow show a relationship between $\displaystyle |x-y|$ and $\displaystyle |z-z_0|$. But I wouldn't know how to do that, either.

Please advise.