Show that there is a neighborhood around $0$ such that $f(z) \neq 0$. The full question is : $f(z)$ is analytic at $z= 0$ and we know that $f(0) \neq 0$. Show that there is a neighborhood around $0$ such that $f(z) \neq 0$.
I have solved this problem before, but when i looked back i do not understand again.

What i do understand is the most intuitive way, since $f(z)$ is
  analytic at $0$, it is by definition analytic around a small
  neighborhood of $0$. In details, take a open ball around $0$, call it
  $B(0,R)$, then $f(z)$ is analytic throughout this open ball. Then,
  since we know that in a analytic domain there can not be infinitely
  many zeroes. Hence there are finite number of zeroes in this region.
  So take the zero with the  minimum distance from the origin, call it
  $r$, form a ball $B(0,r)$, or you can form a ball $B(0,\dfrac{r}{2})$
  then we guarantee there is no zeroes inside.

However, some of my mentors have another more rigorous approach involving the identity theorem, however, they did not elaborate and hence i cannot figure how does identity theorem play a part here. In fact, one of the person on this forum answered me before, but i still cannot figure out: here is his answer
we can ensure that $g(z)$ does not have a zero in the neighbourhood of $0$ that we choose by choosing a small enough neighbourhood.If $g$ had zeros in every neighbourhood (Why is he mentioning every neighborhood?) of $0$, then $0$ would be a limit point of zeros of $g$, and hence we would have that $g(0)=0$ by continuity. (In fact we would have that $g$ is identically $0$, since non-constant analytic functions have isolated zeros.)
 A: You only need the fact that $f$ is continuous at $z=0$, which is a way weaker assumption than the analycity. Write the epsilon-delta definition of continuity at $0$ with $\epsilon = \lvert f(0) \rvert /2$.
In your post, you give two proofs that are moslty equivalent and use the principle of permanence. Both are correct but it is way too complicated to show such a simple fact.
Let me rewrite the second proof :

By contradiction, assume that every neighbourhood of $0$ contains a zero of $f$. Then there is a sequence $(z_n)$ of zeros of $f$ that converges to $0$. Indeed, consider the open ball centered at $0$ with radius $1/n$. It is a neighbourhood of $0$, hence it contains a zero $z_n$. By the principle of permanence, the set of zeros of $f$ has a limit point, which is a contradiction.

A: Assume that for any $n \in \mathbb{N}$ there exists $z_n \in \mathbb{C}$ such that $f(z_n)=0$ and $|z_n| \leq 1/n$. Then $0$ would be a limit point for the set of zeroes of $f$, and this would imply that $f$ is identically zero because non-zero analytic functions have isolated zeros. 
So if $f$ is not the zero function there exists $n$ such that $f(z) \neq 0$ for all $|z| < 1/n$, that is what you want.
