Limit Point is Limit of Convergent Sequence

Theorem
Let $E$ be a subset of metric space $(X,d)$. Let $p$ be a limit point of $E$.

Then, there is a sequence $\{x_n\}$ contained within $E$ such that $x_n \to p$.

Proof 1
We may construct a sequence $\{p_n\}$ using finite recursion.

Let $p_0$ be some arbitrary point in $E$ not equal to $p$.

$p_{n+1}$ is some arbitrary point in $E \cap N_{\frac{d(p,p_n)}{2}}(p)$ where $N_{\frac{d(p,p_n)}{2}}(p)$ denotes the ${\frac{d(p,p_n)}{2}}$-neighborhood of $p$. We can be assured that such a point exists by the definition of a limit point.

Then, $d(p_n,p)<2^{-n}d(p_0,p)$. Since $d(p_0,p)$ is fixed, $d(p_n,p)<2^{-n}d(p_0,p)<r$ for some $n$ because the natural numbers are Archimedean. Therefore, $p$ is the limit of the sequence.

Proof 2
This theorem is a special case of Equivalence of Limit Point Definitions since every metric space induces a topological space.