Let $(X, d)$ be a metric space and suppose that $Y \subseteq X$ is bounded in $(X, d)$then there exists an $x \in X$ and $r > 0$ such that $Y \subseteq B(x, r)$, what is the smallest $r'$ such that $\overline{Y} \subseteq B(x, r')$?
Taking a guess on a bound for $r$, I managed to show that $\overline{Y} \subseteq B(x, 2r)$. If I define $r' = \inf\{ \epsilon> 0 \ | \overline{Y} \subseteq B(x, \epsilon)\}$, then we would have $\overline{Y} \subseteq B(x, r')$, but can I find a nicer "expression" for $r'$in terms of $r$?
I know that taking $r' = r$ doesn't hold, an explicit counterexample is given by $(\mathbb{R}, d)$ where $d$ is the usual metric and $Y = (0,1) = B(\frac{1}{2}, \frac{1}{2})$, letting $r = r' = \frac{1}{2}$ we then see that $\overline{Y} = [0, 1] \not\subseteq B(\frac{1}{2}, r')$
Consider $\Bbb{R}$ with the usual metric, $d(x, y) = |x - y|$. Let $Y = (-1/n, 1/n)$, for some $n \ge 2$ and let $r = 1/2$. Then $Y \subseteq B(0, r)$ and the quantity $r'$ you are looking for is $1/n$. However, you can't calculate $r'$ from $r$ alone: it depends on $n$.