Proof of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Uniformly Continuous

Topological notions on $\mathbb{R}^n$ refer to the topology of the sets open in $(\mathbb{R}^n,d)$, which is a topology by Metric Open Sets Form a Topology.

By claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set there is a real number $R>0$ such that $g(x)=0$ for every $x\in\mathbb{R}^n$ with $\lVert x\rVert>R$. Let $K$ be the closed ball of centre the origin and radius $R+1$ in $(\mathbb{R}^n,d)$; it is compact by claim 2 of A Closed Euclidean Ball is Convex and Compact. By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ the distance from a point $x$ to the origin is $\lVert x\rVert$, so $K$ consists exactly of the points $x$ with $\lVert x\rVert\le R+1$.

The restriction of $g$ to $K$ is continuous on $K$: given $x\in K$ and a real $\varepsilon>0$, any $\delta>0$ witnessing continuity of $g$ at $x$ relative to $\mathbb{R}^n$ also witnesses continuity at $x$ relative to the subset $K$ of $\mathbb{R}^n$. Hence, by Heine-Cantor Theorem: Continuity on a Compact Subset Implies Uniform Continuity, that restriction is uniformly continuous on $K$.

Now let $\varepsilon>0$ be real. Choose $\delta_0>0$ such that all $x,y\in K$ with $d(x,y)<\delta_0$ satisfy $d(g(x),g(y))<\varepsilon$, and let $\delta$ be the smaller of $\delta_0$ and $1$, so that $\delta>0$, $\delta\le\delta_0$ and $\delta\le1$. Let $x,y\in\mathbb{R}^n$ satisfy $d(x,y)<\delta$.

If both $x$ and $y$ lie in $K$, then $d(x,y)<\delta_0$ and the choice of $\delta_0$ gives $d(g(x),g(y))<\varepsilon$.

Otherwise at least one of the two points fails to lie in $K$; since $d$ is symmetric and the desired conclusion is symmetric in $x$ and $y$, we may assume it is $x$, so that $\lVert x\rVert>R+1$. By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ we have $\lVert x-y\rVert=d(x,y)<\delta\le1$, and by claim 6 of the same lemma, applied to the points $y$ and $x-y$ whose sum is $x$,

$$\lVert x\rVert\le\lVert y\rVert+\lVert x-y\rVert<\lVert y\rVert+1 .$$

Hence $\lVert y\rVert>\lVert x\rVert-1>R$, and also $\lVert x\rVert>R+1>R$. By the choice of $R$ both $g(x)$ and $g(y)$ are $0$, so $g(x)=g(y)$ and therefore $d(g(x),g(y))=0<\varepsilon$, because a metric assigns distance $0$ to a point and itself.

In both cases $d(g(x),g(y))<\varepsilon$. As $\varepsilon$ was arbitrary, $g$ is uniformly continuous on $\mathbb{R}^n$.