Proof of A Compact Subset of an Open Set Admits a Uniform Ball Radius

Order arithmetic is that of Elementary Order Arithmetic in an Ordered Field; recall from its preamble that the order of the ordered field $\mathbb{R}$ is a total order, and set $2=1+1$.

Two degenerate cases. Take $r=1$, which satisfies $0<r$ by claim 6 of Elementary Order Arithmetic in an Ordered Field. If $K$ is empty the assertion holds vacuously. If $\Omega=X$ then $\bar B_{d}(x,1)\subseteq X=\Omega$ for every $x\in K$, since every closed ball is a subset of $X$ by Closed Ball in a Metric Space.

So assume from now on that $K$ is nonempty and that the set $A=\{z\in X:z\notin\Omega\}$ is nonempty. Write $\operatorname{dist}_{d}(z,A)$ for the distance from $z$ to $A$ in $(X,d)$, and let $F:K\to\mathbb{R}$ be given by $F(z)=\operatorname{dist}_{d}(z,A)$.

Step 1 (a minimising point). Let $x\in K$ and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Put $\delta=\varepsilon$. Every $y\in K$ with $d(x,y)<\delta$ satisfies, by claim 4 of The Distance to a Set is Nonexpansive and claim 2 of Elementary Order Arithmetic in an Ordered Field,

$$|F(y)-F(x)|\le d(x,y)<\varepsilon .$$

Thus $F$ has the continuity property required in Extreme Value Theorem on a Compact Subset of a Metric Space, and $K$ is nonempty and compact in $X$. That theorem therefore provides $x_{\min}\in K$ with

$$F(x_{\min})\le F(x)\qquad\text{for every }x\in K .$$

Put $m=F(x_{\min})$.

Step 2 ($m$ is positive). Since $x_{\min}\in K\subseteq\Omega$ and $\Omega$ is open in $(X,d)$, Open Subset of a Metric Space provides a real number $s$ with $0<s$ and $B_{d}(x_{\min},s)\subseteq\Omega$, where $B_{d}$ denotes the open ball.

Let $a\in A$. Then $a\notin\Omega$, hence $a\notin B_{d}(x_{\min},s)$, so $d(x_{\min},a)<s$ fails. As $\le$ is a total order, either $s\le d(x_{\min},a)$, or $d(x_{\min},a)\le s$; in the second case $d(x_{\min},a)=s$, since otherwise $d(x_{\min},a)<s$. In both cases $s\le d(x_{\min},a)$.

Hence $s$ is a lower bound for the set $S_{x_{\min},A}=\{t\in\mathbb{R}: t=d(x_{\min},a)\text{ for some }a\in A\}$ of Distance from a Point to a Nonempty Subset of a Metric Space. Since $m=\operatorname{dist}_{d}(x_{\min},A)$ is by that definition the greatest lower bound of $S_{x_{\min},A}$, we get $s\le m$, and therefore $0<m$ by claim 2 of Elementary Order Arithmetic in an Ordered Field.

Step 3 (conclusion). Put $r=m\cdot 2^{-1}$. By claim 8 of Elementary Order Arithmetic in an Ordered Field, $0<r$ and $r<m$.

Let $x\in K$ and let $y\in\bar B_{d}(x,r)$, so that $d(x,y)\le r$ by Closed Ball in a Metric Space. Suppose, for a contradiction, that $y\notin\Omega$; then $y\in A$, so claim 2 of The Distance to a Set is Nonexpansive gives

$$\operatorname{dist}_{d}(x,A)\le d(x,y)\le r .$$

On the other hand $x\in K$, so $m\le F(x)=\operatorname{dist}_{d}(x,A)$ by Step 1. Combining, $m\le r$ by transitivity of the total order $\le$ on $\mathbb{R}$ (Total Order on a Set). Together with $r<m$ this yields $m<m$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, which is impossible because $m<m$ includes $m\ne m$.

Therefore $y\in\Omega$. As $y\in\bar B_{d}(x,r)$ was arbitrary, $\bar B_{d}(x,r)\subseteq\Omega$; and as $x\in K$ was arbitrary, this holds for every $x\in K$. $\blacksquare$