Let $\{X_n\}$ be a bounded sequence.
$a)$ Prove that there exists an $s$ such that for any $r>s$ there exists an $M \in \mathbb{N}$ such that for all $n \geq M$ we have $x_n < r$.
$b)$ If $s$ is a number as in $a)$ then prove $limsup \ X_n \leq s$
$c)$ . Show that if $S$ is the set of all $s$ as in $a)$ then $limsup \ X_n = inf \ S$.
My thoughts: Since $\{X_n\}$ is a bounded sequence there exists a convergent subsequences $\{X_{n_i}\}$ converging to $limsup\ X_n=L$. Therefore for all $\epsilon >0$ there exists an $M \in \mathbb{N}$ such that for if $n \in \mathbb{N}$ and $n \geq M$ then $|X_n - L| < \epsilon$. Now we can define $s=sup\{X_k : k \geq M\}$ then by definition $s$ is the least upper bound for $\{X_n\}$ for all $n \geq M$ and so $|X_n| \leq s < r$ and so $X_n < r$ as required.
Now from here I can't think of how to get $b)$ and $c)$ any help is appreciated, thanks!