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My Proof: Since $f(z)$ is analytic at $z_0$ then it is differentiable at $z_0$

1) $f(z_0)$ exists because $f(z)$ is differentiable at $z_0$

2) $\lim_{z\rightarrow z_0}[f(z)-f(z_0)]=\lim_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{z-z_0}(z-z_0)=\lim_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{z-z_0}\lim_{z\rightarrow z_0}(z-z_0)=f(z_0)\lim_{z\rightarrow z_0}(z-z_0)=0$

3) $\lim_{z\rightarrow z_0}f(z)=f(z_0)$

therefore $f(z)$ is continous at $z_0$.

Is this proof correct?

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    That $f(z_0)$ exists is part of the definition of $f$. The second line isn't really necessary, but if you want to emphasize that $f(z_0)$ is a well defined number, it makes more sense to say something like "Since $f$ is given as a convergent power series in a neighborhood of $z_0$, $z_0$ is in its domain of definition, so $f(z_0)$ is a well defined number". – 2017-01-07

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It is, but because the limit that you wrote is a product of limits that exist and whose product makes sense. In case you want to avoid this, a somewhat cleaner alternative is the following.

Given $\varepsilon>0$, there exists $\delta>0$ such that $$ \left|\frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)\right|<\varepsilon\quad\text{when } |z-z_0|<\delta. $$ Hence, $$ \left|\frac{f(z)-f(z_0)}{z-z_0}\right|<|f'(z_0)|+\varepsilon $$ and we obtain $$ |f(z)-f(z_0)|=\left|\frac{f(z)-f(z_0)}{z-z_0}\right|\cdot|z-z_0|<(|f'(z_0)|+\varepsilon)|z-z_0|. $$ Now just let $z\to z_0$.

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    Is the first point of my proof correct? $f(z_0)$ exists because $f(z)$ is differentiable at $z_0$ – 2017-01-07
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    Do you mean "$f(z_0)$ exists"? If so, yes, we only define the derivative in this case (at this level). – 2017-01-07