Proof of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set

Write $Z=\{x\in\mathbb{R}^n:f(x)\ne0\}$ and $S=\operatorname{supp}f$, so that $S$ is the closure of $Z$ by Support of a Real-Valued Function on a Topological Space; by The Closure is the Smallest Closed Superset the set $S$ is closed, contains $Z$, and is contained in every closed set that contains $Z$. Throughout we use that $d(x,0)=\lVert x\rVert$ for every $x\in\mathbb{R}^n$: indeed $d(x,0)=\lVert x-0\rVert$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, and $x-0=x$ in the real vector space $\mathbb{R}^n$.

Claim 1. If $x\notin S$ then $x\notin Z$, because $Z\subseteq S$; that is, $f(x)=0$.

Claim 3. Let $R>0$ be a real number such that $f(x)=0$ for every $x$ with $\lVert x\rVert>R$, and let

$$\bar B=\{x\in\mathbb{R}^n: d(x,0)\le R\}$$

be the closed ball of centre $0$ and radius $R$. If $x\in Z$ then $f(x)\ne0$, so $\lVert x\rVert>R$ is impossible; hence $d(x,0)=\lVert x\rVert\le R$ and $x\in\bar B$. Thus $Z\subseteq\bar B$. By A Closed Euclidean Ball is Convex and Compact the set $\bar B$ is compact, and therefore closed by Compact Subset of $\mathbb{R}^n$ is Closed. Being a closed set containing $Z$, it contains $S$, which is claim 3.

Claim 2. Suppose first that a real number $R>0$ as in claim 2 exists. By claim 3 we have $S\subseteq\bar B$ and hence $S=S\cap\bar B$. Since $S$ is closed and $\bar B$ is compact, the intersection $S\cap\bar B$ is compact by Compactness of Intersections with Closed Sets and of Level Sets of Semicontinuous Functions. So $S$ is compact and $f$ is compactly supported.

Conversely, suppose $S$ is compact. By Compact Subset of $\mathbb{R}^n$ is Bounded it is bounded, so there are a point $x_0\in\mathbb{R}^n$ and a real number $R_0>0$ with $d(x_0,y)\le R_0$ for every $y\in S$. Put

$$R=d(x_0,0)+R_0+1,$$

which is a real number with $R>0$. For $y\in S$, the symmetry and the triangle inequality of the metric $d$ give

$$\lVert y\rVert=d(y,0)\le d(y,x_0)+d(x_0,0)=d(x_0,y)+d(x_0,0)\le R_0+d(x_0,0)<R.$$

Consequently no $x$ with $\lVert x\rVert>R$ lies in $S$, and for every such $x$ claim 1 gives $f(x)=0$. This is the stated condition, and claim 2 is proved.