Let $(X,d)$ be a metric space. Assume that $(x_n)$ is a sequence in $X$. There is a well-known theorem about the convergence of $(x_n)$ that reads as follows.
Theorem. Every sub-sequence of $(x_n)$ has a sub-sub-sequence of $(x_n)$ that convergence to $x$ if and only if the sequence $(x_n)$ converges to $x$.
The proof for $\impliedby$ is easy and immediate. For $\implies$, there are proofs in this post which are using contradiction. I was wondering if there is a direct proof for $\implies$?
It is not too difficult to adapt the proof in the case of the real numbers that was linked to in the comments to work in an arbitrary metric space.
The key is to recall that $x_n \to x$ in $X$ if and only if $d(x_n,x) \to 0$ in $\mathbb{R}$.
Now it is clear that $\liminf_{n \to \infty} d(x_n,x) = 0$ since $x_n$ certainly has a subsequence converging to $x$. Additionally, there is a subsequence $x_{n_k}$ such that $d(x_{n_k}, x) \to \limsup_{n \to \infty} d(x_n,x)$ as $k \to \infty$. By passing to a further subsequence, we can assume that $x_{n_k} \to x$ in $X$. This means that $d(x_{n_k},x) \to 0$ so that $\limsup_{n \to \infty} d(x_n,x) = 0$.
Therefore $\lim_{n \to \infty} d(x_n,x) = 0$ as desired.
Here's an elementary argument that holds whenever $ X $ is a topological space.
For any subset $ J \subset \mathbb N $, we define $ X_J = \{ x_n : n \in J \} $. Given an infinite subset $ J \subset \mathbb N $, we can identify a subsequence $ ( x_{ n_k } ) $ of $ ( x_n ) $ with each $ n_k $ coming from $ J $. In particular, the subsequence $ ( x_{ n_k } ) $ lies in $ X_J $. Applying the hypothesis, there exists a further subsequence $ ( x_{ n_{ k_l }} ) $ converging to $ x $, from which we conclude that $ x $ lies in $ \overline{ X_J} $, the closure of $ X_J $. This establishes that $$ J \text{ infinite} \quad \Longrightarrow \quad x \in \overline{ X_J} . $$
Fix now a neighborhood $ U $ of $ x $ and consider the set $ J = \{ n \in \mathbb N : x_n \notin U \} $. Since $ X_J = \{ x_n : x_n \notin U \} $, the elements of $ X_J $ lie outside of $ U $, so they cannot get close to $ x $. More precisely, $ x $ does not lie in $ \overline{ X_J } $, and by the above, the set $ J $ must be finite. Consequently, every neighborhood $ U $ of $ x $ contains all but finitely many terms of the sequence, or the sequence $ ( x_n ) $ converges to $ x $.
