Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit

Theorem
Let $\left({X, \tau}\right)$ be a first-countable topological space.

Let $\left\langle{x_n}\right\rangle_{n \in \N}$ be an infinite sequence in $X$.

Let $x$ be an accumulation point of $\left\langle{x_n}\right\rangle$.

Then $x$ is a subsequential limit of $\left\langle{x_n}\right\rangle$.

Proof
By the definition of a first-countable space, there exists a countable local basis $\mathcal B$ at $x$.

By Surjection from Natural Numbers iff Countable, there exists a surjection $\phi: \N \to \mathcal B$.

For all $n \in \N$, define the set:
 * $\displaystyle U_n = \bigcap_{k \mathop = 0}^n \phi \left({k}\right)$

By General Intersection Property of Topological Space, it follows that $U_n$ is an open neighborhood of $x$.

Using the principle of recursive definition, we construct a strictly increasing sequence $\left\langle{n_k}\right\rangle_{k \in \N}$ in $\N$.

By the definition of an accumulation point, we can choose $n_0 \in \N$ such that $x_{n_0} \in U_0$.

For all $k \in \N$, let $n_{k+1} > n_k$ be the (unique) smallest natural number such that $x_{n_{k+1}} \in U_{k+1}$.

Such an $n_{k+1}$ exists by the definition of an accumulation point, and by the well-ordering principle.

We now show that $x$ is a limit point of $\left\langle{x_{n_k}}\right\rangle$.

Let $U$ be an open neighborhood of $x$.

By the definition of a local basis, there exists an $H \in \mathcal B$ such that $H \subseteq U$.

By the definition of a surjection, there exists a natural number $m$ such that $H = \phi \left({m}\right)$.

By construction, we have:
 * $\forall k \in \N: k > m \implies x_{n_k} \in U_k \subseteq \phi \left({m}\right) = H$

Hence the result, by the definition of a limit point.