Proof of Uniform Continuity Along a Compact Subset of the Domain

Order arithmetic is that of Elementary Order Arithmetic in an Ordered Field, absolute values have the properties of Properties of the Absolute Value in an Ordered Field, and $2=1+1$. Note that $d_{\mathbb{R}}(f(z),f(x))=|f(z)-f(x)|$ throughout.

If $K$ is empty, take $\theta=1$, which is positive by claim 6 of Elementary Order Arithmetic in an Ordered Field; the assertion holds vacuously. So assume $K$ is nonempty.

Let $\eta\in\mathbb{R}$ with $0<\eta$ and put $\eta'=\eta\cdot 2^{-1}$, so that $0<\eta'$ and $\eta'+\eta'=\eta$ by claim 8 of Elementary Order Arithmetic in an Ordered Field.

Step 1 (a cover indexed by pairs). For $x\in K$ let

$$G_{x}=\{\tau\in\mathbb{R}: 0<\tau\text{, and every }z\in\Omega\text{ with }d(x,z)<\tau\text{ satisfies }|f(z)-f(x)|<\eta'\}.$$

Since $x\in K\subseteq\Omega$ and $f$ is continuous at $x$ relative to $\Omega$, that definition applied with $\eta'$ provides an element of $G_{x}$; hence $G_{x}$ is nonempty.

Let $I=\{(x,\tau): x\in K\text{ and }\tau\in G_{x}\}$, and for $i=(x,\tau)\in I$ put $W_{i}=B_{d}(x,\tau\cdot 2^{-1})$, the open ball of centre $x$ and radius $\tau\cdot 2^{-1}$, which is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field. Then $(W_{i})_{i\in I}$ is a family of subsets of $X$, each $W_{i}$ is open in $(X,d)$ by Open Ball in a Metric Space is Open, and every $x\in K$ lies in $W_{(x,\tau)}$ for any $\tau\in G_{x}$, because $d(x,x)=0<\tau\cdot 2^{-1}$ by condition 2 of Metric Space. Hence $(W_{i})_{i\in I}$ is an open cover of $K$ in $X$.

Step 2 (a finite subcover and a uniform radius). Since $K$ is compact in $X$, Compact Subset Criterion via Open Covers in the Ambient Space provides a finite subset $J\subseteq I$ with $K\subseteq\bigcup_{j\in J}W_{j}$. As $K$ is nonempty and a union indexed by the empty set is empty, $J$ is nonempty. By Finite Set the set $J$ therefore has $n$ elements for some natural number $n$; fix a bijection $\beta:[n]\to J$ from the initial segment $[n]$ onto $J$, and for $k\in[n]$ write $\beta(k)=(w_{k},\tau_{k})$.

Let $c$ be the $n$-tuple in $\mathbb{R}$ with components $c_{k}=-(\tau_{k}\cdot 2^{-1})$. By Greatest Element of a Finite Family in a Totally Ordered Set, applied to the totally ordered set $\mathbb{R}$, there is $j\in[n]$ with $c_{k}\le c_{j}$ for every $k\in[n]$; by claim 4 of Elementary Order Arithmetic in an Ordered Field this says

$$\tau_{j}\cdot 2^{-1}\le\tau_{k}\cdot 2^{-1}\qquad\text{for every }k\in[n].$$

Put $\theta=\tau_{j}\cdot 2^{-1}$; then $0<\theta$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, and $\theta\le\tau\cdot 2^{-1}$ for every pair $(w,\tau)\in J$, since $\beta$ is onto $J$.

Step 3 (the estimate). Let $x\in K$ and let $z\in\Omega$ satisfy $d(x,z)<\theta$. By Step 2 there is $j\in J$ with $x\in W_{j}$; write $j=(w,\tau)$, so that $w\in K$, $\tau\in G_{w}$ and $d(w,x)<\tau\cdot 2^{-1}$.

By condition 4 of Metric Space and claims 3 and 2 of Elementary Order Arithmetic in an Ordered Field,

$$d(w,z)\le d(w,x)+d(x,z)<\tau\cdot 2^{-1}+\theta\le\tau\cdot 2^{-1}+\tau\cdot 2^{-1}=\tau ,$$

the last equality by claim 8 of Elementary Order Arithmetic in an Ordered Field. Since $z\in\Omega$ and $\tau\in G_{w}$, this gives $|f(z)-f(w)|<\eta'$.

Likewise $d(w,x)<\tau\cdot 2^{-1}<\tau$ by claim 8, so $d(w,x)<\tau$ by claim 2 of Elementary Order Arithmetic in an Ordered Field; and $x\in K\subseteq\Omega$, so $|f(x)-f(w)|<\eta'$.

Finally, by claims 5 and 2 of Properties of the Absolute Value in an Ordered Field and claim 3 of Elementary Order Arithmetic in an Ordered Field,

$$|f(z)-f(x)|=\bigl|\bigl(f(z)-f(w)\bigr)+\bigl(f(w)-f(x)\bigr)\bigr|\le|f(z)-f(w)|+|f(w)-f(x)|<\eta'+\eta'=\eta .$$

Since $x\in K$ and $z\in\Omega$ with $d(x,z)<\theta$ were arbitrary, $\theta$ has the required property. $\blacksquare$