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 $\implies$?
