Request:
Prove that $x$ is an accumulation point of a set $S$ iff there exists a sequence $(s_n)$ of points in $S\setminus \{x\}$ such that $(s_n)$ converges to $x$.
Attempt:
Since this is a show-me-what-you've-got-so-far community, I'll give you what I've come up with so far even though it's incomplete:
Suppose that $x$ is an accumulation point of a set $S$. By the definition of accumulation point, for all $\epsilon>0$ we have that $N^*(x;\epsilon)\cap S\neq \varnothing$. Thus, we can conclude that $N^*(x;\frac{1}{n})\cap S\neq \varnothing$ by setting $\epsilon_n=\frac{1}{n}$ for $n\in\mathbb{N}$, and so for all $n\in\mathbb{N}$ we can select the desired sequence $s_n\in N^*(x;\frac{1}{n})\cap S$. Now, for all $n\in\mathbb{N}$ we have $s_n\neq x$, and $s_n\in S$ since $s_n\in N^*(x;\frac{1}{n})\cap S$—for all $n\in\mathbb{N}$. To show that for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that if $n>N$, then $\lvert s_n-s\rvert <\epsilon$ let $\epsilon>0$; by the Archimedian property there exists $N\in\mathbb{N}$ such that $\frac{1}{N}<\epsilon$, and by design $s_n\in N^*(x;\frac{1}{n})$ for all $n$. Thus $\lvert s_n-x\rvert <\frac{1}{n}$ for all $n$, and so $n>N$ implies $\lvert s_n-x\rvert <\frac{1}{n}<\frac{1}{N}<\epsilon$ . . . still working . . .
Best Answer
Sequence implies accumulation
Let $X$ be an arbitrary topological space.
Let $S\subseteq X$ and let $x\in S$.
Let $(s_n)$ be a sequence in $S\setminus\{x\}$ converging to $x$.
Let $U$ be an open set containing $x$. Then by the definition of convergence, there is an $N$ such that whenever $n \ge N$, $s_n \in U$. In particular, $s_N\in U$. Since $(s_n)$ is a sequence in $S\setminus \{x\}$, $s_N \in S\cap U$ and $s_N \ne x$. That is, every open neighborhood of $x$ contains an element of $S$ not equal to $x$, so $x$ is an accumulation point of $S$.
Accumulation implies sequence in a metric space:
Let $(X,d)$ be a metric space, let $S\subseteq X$, and let $x$ be an accumulation point of $S$.
For each positive integer $n$, let $B_n$ be the open ball about $x$ with radius $\frac1n$. Since $x$ is an accumulation point of $S$, $S\cap B\setminus\{x\}$ is non-empty for each $n$. Thus by the axiom of countable choice, there is a sequence $(s_n)$ in $S$ such that $s_n\in B_n$ for each $n$. Then since the set of all these open balls form a neighborhood basis at $x$, $(s_n)$ converges to $x$.
Accumulation does not imply sequence in a general topological space:
Let $S=\omega_1$ be the first uncountable ordinal. Let $X=\omega_1^+$ be considered under the order topology. Then $\omega_1$ is clearly an accumulation point of $S$. Suppose $(s_n)$ is a sequence in $S$. Then $(s_n)$ is bounded above by its union, which is a countable union of countable ordinals, and hence a countable ordinal. Thus $\bigcup_n s_n<\omega_1$, but $s_n$ never exceeds it, so $(s_n)$ does not converge to $\omega_1$.
Edit: The fact that a countable union of countable sets is countable is called the countable union condition, and is a form of the axiom of choice weaker than the axiom of countable choice but stronger than the axiom of countable choice for collections of finite sets.