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

We work in the ordered field of real numbers, whose order is in particular a total order, and we write $u-v$ for $u+(-v)$.

Since $A$ is bounded in $(X,d)$, there are a point $x\in X$ and a real number $R>0$ such that $d(x,y)\le R$ for every $y\in A$. Fix such $x$ and $R$, and let $z\in\operatorname{cl}_X(A)$.

Suppose, for contradiction, that $d(x,z)\le R$ fails. Since the order is total, this forces $R\le d(x,z)$ and $R\neq d(x,z)$, that is, $R<d(x,z)$. Put $\varepsilon=d(x,z)-R$. By claim 1 of Elementary Order Arithmetic in an Ordered Field, applied with the summand $-R$, the inequality $R<d(x,z)$ gives $R-R<d(x,z)-R$, and $R-R=0$ by claim 3 of Additive Cancellation and Elementary Additive Identities in a Field; hence $0<\varepsilon$.

Applying condition 3 of Characterization of the Closure in a Metric Space by Open Balls at the point $z$, which lies in $\operatorname{cl}_X(A)$, with this $\varepsilon$, we obtain a point $a\in A$ with $d(z,a)<\varepsilon$. By condition 3 of the definition of a metric we have $d(a,z)=d(z,a)<\varepsilon$, and by condition 4 of that definition

$$d(x,z)\le d(x,a)+d(a,z).$$

Since $a\in A$ we have $d(x,a)\le R$. By claim 3 of Elementary Order Arithmetic in an Ordered Field, applied to the strict inequality $d(a,z)<\varepsilon$ and the inequality $d(x,a)\le R$, we get $d(a,z)+d(x,a)<\varepsilon+R$, and by commutativity of addition this reads $d(x,a)+d(a,z)<R+\varepsilon$. Combining with the displayed inequality by claim 2 of Elementary Order Arithmetic in an Ordered Field gives $d(x,z)<R+\varepsilon$.

Finally, by commutativity and associativity of addition and by claim 3 of Additive Cancellation and Elementary Additive Identities in a Field,

$$R+\varepsilon=R+\bigl(d(x,z)+(-R)\bigr)=d(x,z)+\bigl(R+(-R)\bigr)=d(x,z)+0=d(x,z).$$

Hence $d(x,z)<d(x,z)$, which contradicts the fact that $u<v$ requires $u\neq v$. Therefore $d(x,z)\le R$.

Since $z\in\operatorname{cl}_X(A)$ was arbitrary, we have $d(x,z)\le R$ for every $z\in\operatorname{cl}_X(A)$, with $x\in X$ and $R>0$; by the definition of boundedness, $\operatorname{cl}_X(A)$ is bounded in $(X,d)$.