Let $G$ be a connected open set and let $f: G \to \mathbb{C}$ be an analytic function. Then the following are equivalent statements:
a) $f \equiv 0$
b) There is a point $a \in G$ such that $f^{n}(a) = 0$ for each $n > \geq 0$. This statement means there exists a point $a \in G$ such that $a$ is a zero of $f$ with $\infty$ multiplicity.
c)$\{z \in G : f(z) =0\}$ has a limit point in $G$. This statement means the sequence $\{z_n\}$ in $G$ converges to a limit in $G$ and in addition we have $f(z_n) = 0$ for all $n \in \mathbb{N}$.
For this question I prove $c \implies b$, following closely with conway's proof and I will highlight some part that I do not get it, it might be really simple but I lack a bit of metric spaces knowledge. Some of the proof has been elaborated for my better understanding.
Hypothesis says that there exists limit point $a \in G$ of the set $Z = \{z \in G : f(z) =0\}$ which is a closed set in $G$. So this implies $a$ lies in $Z$ as well, hence $f(a) = 0$. I had a little trouble understanding this part. Intuitively, a sequence of $z_n$ goes to a limit point $a$, so $f(a)$ should behave similarly like $f(z_n) = 0$. But I wish to have a more rigorous view.
Claim: $a \in G$ is zero of $\infty$ multiplicity (which if we can show then $(b)$ is proven).
Suppose not, that means there exists a integer $n \geq 1$ such that $f(a) = f^{'}(a) = ... = f^{n-1}(a) = 0$ and $f^{n}(a) \neq 0$. So it means $a$ has zero of order $n$.
Note that $f$ is given to be analytic and of course we can say it is also analytic in $|z- a| < R$. It follows by the Theorem Zeroes and Poles that $f(z) = (z-a)^ng(z)$ where $g(a) \neq 0$.
Now we know $g$ is analytic also and hence $g$ is continuous, and in particular, continuous in $B(a,R)$. By Theorem Neighborhood of non zeros, then there exists $0 < r < R$ such that for all $z \in B(a,r)$, $g(z) \neq 0$.
But since $a$ is a limit point of $Z$ there exists a point $b$ with $f(b) = 0$ and in particular $0 < |b-a| < r$. This gives $0 = (b-a)^ng(b) \implies g(b) = 0$. This bolded part I do not quite understand. As usual, intuitively seems correct but I need some justification.
This contradicts that $g(z) \neq 0$ in the $B(a,r)$. Hence concluding the proof.
Since $a$ is a limit point of $f^{-1}\{0\}$, every neighbourhood of $a$ contains a point $b\in f^{-1}\{0\}$ different from $a$. Therefore, from $$f(b)=(b-a)^ng(b),\enspace f(b)=0\enspace\hbox{and}\enspace b\neq a,$$ you can deduce that $g(b)=0$.