Edited after Jjagmath comment:
I am trying to understand why can we always construct a subsequence that will converge to the limsup.
Rudin defines the $\limsup$ as the supremum of the set of subsequential limits.
I've seen the $\limsup$ also referred to as the supremum of the set of limit points ( proof attempt below)
This seems to suggest that every limit point is a subsequential limit but the converse is false e.g. take $(-1)^{n}$ in $\mathbb{R}$, Then -1 is a subsequential limit but not a limit point? So does that mean the set of limit points is a subset of subsequential limits?
Attempt at trying to show limsup is sup of limit points:
Let $x_{n}$ be a bounded sequence and $b_{n}:= \sup\{a_{k}: k \geq \}$ then by definition $\lim_{n \to \infty}b_{n} = \limsup_{n \to \infty}a_{n}$. Now since $(b_{n})$ converges every subsequence converges to the same limit. So let $(a_{n_{k}})$ be a convergent subsequence. Then $a_{n_{k}} \leq b_{n_{k}} \forall k \in \mathbb{N}$. If I then take limits I get $\lim_{k \to \infty}a_{n_{k}} \leq \lim_{k \to \infty}b_{n_{k}} = \limsup_{n \to \infty} a_{n}$
I am not sure how to then proceed...
Attempt at trying to show we can construct a subsequence that will converge to limsup:
Let $\{x_{n}\}$ be a sequence of real numbers and $\alpha = \limsup x_{n}$ If I assume that limsup is the supremum of all limitpoints then there is some $x_{n_{1}}$ (say) such that $|x_{n_{1}} - \alpha| < \frac{1}{2}$. Similarly for all $k \in \mathbb{N}$, some $k_{n_{k}}$ such that $|x_{n_{k}} - \alpha| < \frac{1}{2^{k}}$. Result follows.