Solution: Let $\alpha $ be the supremum of $A$. We need to show that $\alpha \in \bar{A} $. Suppose not, then there exists $\varepsilon >0$ such that $(\alpha -\varepsilon , \alpha + \varepsilon ) \cap A = \emptyset $. This means, the interval $(\alpha -\varepsilon , \alpha +\varepsilon )$ does not contain any element of $A$ thus for any $a\in A$, \[ a \lt \alpha - \varepsilon \lt \alpha. \] This is a contradiction as $\alpha = \sup A$. Thus, $\alpha \in A$.