I am not too sure how to go about this, below is what I know so far.
Proof: Since $S$ is bounded, $S$ is bounded below because $x=\inf(S)$ exists.
$x=\inf(S)$ implies
1) $\forall y\in S$, $y \ge x$
2) $\forall \varepsilon >0, \exists r\in S$ such that $r<x+\varepsilon$
I also know the definition of a boundary point, but I don't know how to use these definitions to solve the problem.
For $\epsilon>0$, since $x+\epsilon$ is not a lower bound for $S$, there is an $x_{\epsilon}\in S$ such that $x_{\epsilon}<x+\epsilon$, so $0\leq x_{\epsilon}-x<\epsilon$ (actually it is $x_{\epsilon}-x>0$), so $|x_{\epsilon}-x|<\epsilon$, and hence $x_{\epsilon}\in B_{\epsilon}(x)$, so $B_{\epsilon}(x)\cap S$. Now $x\notin S$, so $B_{\epsilon}(x)\cap S^{c}\ne\emptyset$. We conclude that $x\in\overline{S}\cap\overline{S^{c}}=\partial S$.