Let $ (x_n) $ be a sequence of reals.

$ p \in \mathbb{R} $ is an $\color{goldenrod}{\text{accumulation point}}$ of the sequence if for every $ \epsilon > 0 $, $ x_n $ lies in $ (p-\epsilon, p+\epsilon) $ for infinitely many $ n $ (stricten “infinitely many” to “all but finitely many”, and we get definition of limit of a sequence).

A subsequence of $ (x_n) $ is just a sequence $ x_{n_1}, x_{n_2}, \ldots $ with $ n_1 < n_2 < \ldots $.

Notice $ \lbrace \text{limits of convergent subsequences of }(x_n) \rbrace $ $ = \lbrace \text{accumulation points of }(x_n)\rbrace $

A subsequential limit is clearly an accumulation point of seq.
Let $ p \in \mathbb{R} $ be an accumulation point of seq. There is an $ n_1 $ with $ |x_{n_1} - p| < 1 $. Now as $ x_n $ lies in $ (p-\frac{1}{2}, p+\frac{1}{2}) $ for infinitely many $ n $, we can pick an $ n_2 > n_1 $ with $ |x_{n_2} - p| < \frac{1}{2} $. And now we can pick an $ n_3 > n_2 $ with $ | x_{n_3} - p| < \frac{1}{3} $, and so on. This gives a subsequence $ (x_{n_k}) $ converging to $ p $.