$S$ is a bounded nonempty subset of $\mathbb{R}$ such that $\sup S$ is not in $S$. Prove there is a sequence $s_n$ in $S$ such that $\lim s_n = \sup S$.
Proof:
Any sequence $s_n$ where $\sup s - s_n < 1/2^n$ for all $n$ clearly converges to $\sup S$.
Such a sequence exists with all $s_n$ in $S$, as follows: Given any $\epsilon > 0$, there exists $s$ in $S$ such that $\sup S - s < \epsilon$. If not, $\sup S - \epsilon$ would be an upper bound on $S$.