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?
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 $z\to z_0$.