I asked a question earlier today, but I still feel like there is room for another question here.
For the following, $X$ is a metric space, $A\subseteq X$, and the notation $\lbrace a_n \rbrace_{n\in \mathbb{N}}\rightarrow x$ denotes the sequence converging to the limit $x\in X$.
$\textbf{(1)}$ $\forall$ sequence $\lbrace a_n \rbrace_{n\in \mathbb{N}}$ that is a subset of $A$, [if $\lbrace a_n \rbrace_{n\in \mathbb{N}}\rightarrow x$, then $x\in A$].
$\textbf{(2)}$ the set $A$ is closed (i.e. $\forall \text{ limit point } c \text{ of } A, c\in A.$).
$\textbf{Proof for why (1) implies (2):}$ Assume $\forall$ sequence $\lbrace a_n \rbrace_{n\in \mathbb{N}}$ that is a subset of $A$, [if $\lbrace a_n \rbrace_{n\in \mathbb{N}}\rightarrow x$, then $x\in A$]. Now, let $c$ be an arbitrary limit point of $A$. Thus, it is enough to show $c\in A$ to complete the proof. From the definition of a limit point, it is true that $\forall r >0, (B_r(c)-\lbrace c\rbrace)\cap A\neq \phi$. Hence, $\forall n\in \mathbb{N}, (B_\frac{1}{n}(c)-\lbrace c\rbrace)\cap A\neq \phi$. Thus, $(\exists \lbrace x_i \rbrace_{i\in \mathbb{N}}\subseteq A)( \forall n\in \mathbb{N}) [x_n\in (B_\frac{1}{n}(c)-\lbrace c\rbrace)]$. Say $\lbrace a_i \rbrace_{i\in \mathbb{N}}$ is such a sequence that satisfies this argument. Thus, by the original assumption, it is enough to show $\lbrace a_i \rbrace_{i\in \mathbb{N}}\rightarrow c$ to figure $c\in A$ to complete the proof.
$\textbf{Question}:$ Why/how does $\lbrace a_i \rbrace_{i\in \mathbb{N}}\rightarrow c$?
Best Answer
When you say that $\{a_i\}_{i\in\mathbb{N}}$ is a sequence that satisfies this argument you are saying that $a_n\in B_{\frac{1}{n}}(c)-\{c\}$, so $d(a_n,c)<\frac{1}{n}$ and convergence follows.
The outline of the proof for $(1)\implies (2)$ is to construct a sequence using the definition of limit point which, by construction, converges to the limit point. Then, the assumption $(1)$ gives that the limit point is contained in the set, and we have $(2)$.