Prove that limsup is the supremum of the limit points

Let ${a_n}$ be a bounded sequence of real numbers, and let P be the set of limit points. Prove that $\limsup a_n = \sup P$.

I have proved that there must be a subsequence that converges to limsup$a_n$. But I am stuck connecting this to the maximum limit point!

Let $b_n:=\sup\{a_k:k\ge n\}$. By definition $\lim_{n\to \infty}b_n=\limsup_{n\to \infty}a_n$. Note that since $(b_n)$ converges, every subsequence converges to the same limit. Now let $(a_{n_k})$ be a convergent subsequence. Then $a_{n_k}\le b_{n_k}$ for all $k\in \mathbb{N}$. It follows that $\lim_{k\to \infty}a_{n_k}\le \lim_{k\to \infty}b_{n_k}=\limsup_{n\to \infty}a_n$.

So far, you have shown that there is a subsequence converging to $\limsup a_n$. I.e., you have shown $\limsup a_n \in P$, which implies $$ \limsup a_n \le \sup P $$ To show the other inequality, consider any value $x \in P$. Then there is a subsequence of $a_n$, call it $a_{n_i}$, converging to $x$. Then notice that $$ a_{n_i} \le \sup_{n > n_i} a_n $$ So taking the limit of the above inequality (which exists for both sides), we find $$ x \le \limsup a_n $$ And since $x \in P$ was arbitrary, $$ \sup P \le \limsup a_n. $$