The Closure of a Bounded Subset of a Metric Space is Bounded

Let $(X,d)$ be a metric space, and let $\mathcal{T}_d$ be the collection of all subsets of $X$ that are open in $(X,d)$, which is a topology on $X$ by Metric Open Sets Form a Topology. Let $A\subseteq X$ be bounded in $(X,d)$.

Then the closure $\operatorname{cl}_X(A)$ of $A$ in $(X,\mathcal{T}_d)$ is bounded in $(X,d)$. More precisely, if $x\in X$ and $R>0$ are such that $d(x,y)\le R$ for every $y\in A$, then $d(x,z)\le R$ for every $z\in\operatorname{cl}_X(A)$.