The full question is:
Suppose $u = sup(S)$, where $S$ is a nonempty bounded set in $\mathbb R$. Show that there is a sequence $s_n$ in $S$ such that $s_n \rightarrow u$.
I was wondering if I simply say that any sequence $s_n$ in $S$ that contains $\max \{S\}$ converges to $u$ by the monotonic convergence theorem by definition.
The reasoning being that since any bounded, monotonic increasing sequence converges to the supremum of the sequence, $s_n$, which is contained within a set with the supremum of $u$, would also converge to $u$.
Does that work, or do I need to be more rigorous?
By definition of the supremum of S, for every $\epsilon>0$ there is a point $x\in S$ such that $$\sup(S)-\epsilon<x.$$ This implies that for every $n\in\mathbb N$ there is a point $x_n\in S$ such that $$x_n>\sup(S)-\frac 1 n.$$ This inequality shows that $x_n\to \sup(S).$