Here is my proof. (Note that $z$ refers to complex number.)
Since $f(z)$ is continuous at $z=z_0$, by definition, for any $\epsilon > 0$, there exists $\delta > 0$ such that $|f(z)-f(z_0)| <\epsilon$ whenever $|z-z_0|<\delta$. (or equivalently, there exists a neighbourhood $N(z_0;\delta)$.)
Then for $\epsilon=|f(z_0)|>0$ ($|f(z_0)|\neq 0$ as $f(z_0)\neq 0$), there exists $N(z_0;\delta)$ such that
$$\big||f(z)|-|f(z_0)|\big|\leq |f(z)-f(z_0)|<\epsilon$$ $$\Rightarrow\big||f(z)|-|f(z_0)|\big|<\epsilon$$ $$\Rightarrow -\epsilon+|f(z_0)|<|f(z)|<\epsilon+|f(z_0)|$$ $$\Rightarrow 0<|f(z)|<2|f(z_0)|$$ $$\Rightarrow |f(z)|>0$$ $$\Rightarrow f(z)\neq 0$$ as required.
Is my proof correct?
Assume the contradiction and in the definition of continuity, take $\epsilon=\alpha |P(z_0)|$ for a fixed $\alpha$ with $|\alpha|<1.$ Since $f(z)$ vanishes for some value of $z$ in every nhd of $f(z_0)$ we will have a situation $|f(z_0)|<|\alpha f(z_0)|,$ a contradiction.