Why must bounded sets be contained within a closed ball?

The definition of a bounded set $S$ in a metric space $X$, is simply that $$S\subset B_r(x)$$ for some $r>0$ and $x\in X$. But I've seen a few proofs where they use the fact that a set is bounded to say that it must be contained in some closed ball. I can't think of an example where this isnt true, but I wasn't sure if the two definitions are equivalent. Can someone give a proof? Thanks!

• Every open ball is contained in a closed ball. Every closed ball is contained in an open ball. Jun 12 '17 at 5:58

If $S$ is bounded then there exist $x\in X,r>0$ such that $S\subseteq B_r(x)$. But $B_r(x)\subseteq \overline{B_t}(x)$, where $t$ is any number you like with $t>r$.