Suppose a set $S$ of real numbers is bounded and let $\mu$ be an upper bound for $S$. Show that $\mu$ is the least upper bound of $S$ $\Longleftrightarrow$ for every $\epsilon > 0$ there is an element of $S$ in the interval $[\mu - \epsilon, \mu]$.
My Work
($\Rightarrow$)
If there is no element of $S$ in the interval $[\mu - \epsilon, \mu]$, then $\mu - \epsilon$ could also be an upper bound for $S$, but since $\mu = \sup S$ and $\mu - \epsilon < \mu$ there is a contradiction.
($\Leftarrow$)
Considering the least upper bound of $S$, I need to show that it cannot be smaller than $\mu$. But I am not sure how to do this using the condition I am given.
Edit
By the definition of a supremum, if $\lambda$ is another upper bound of $S$, then if $\mu = \sup S \Rightarrow \mu \le \lambda$. So proof by contradiction, assuming that $\mu \ne \sup S$, does that mean that there is an element $\lambda \in [\mu - \epsilon, \mu], \lambda \notin S, \lambda < \mu$? How does this prove that there is no element of $S$ in $[\mu - \epsilon, \mu]$ for every $\epsilon > 0$?