The function equal to $e^{-1/x^2}$ for $x>0$ and $0$ for $x\le0$ is $C^\infty$, and composing it with $1-\abs{x}^2$ gives a nonzero $C^\infty$ function with compact support.Suggest a slogan
Let \(f\colon\RR\to\RR\) be given by \(f(x)\coloneqq e^{-1/x^2}\) for \(x>0\) and \(f(x)\coloneqq0\) for \(x\leq0\). Then \(f\in C^\infty(\RR)\), and \(\psi(x)\coloneqq f(x)\,f(1-x)\) is a \(C^\infty\) function on \(\RR\) with \(\psi>0\) on \((0,1)\) and \(\psi=0\) outside \((0,1)\). For \(n\geq1\), \(\phi(x)\coloneqq f\big(1-\abs{x}^2\big)\) is a \(C^\infty\) function on \(\RR^n\) with \(\supp\phi=\theset{x : \abs{x}\leq1}\) [@Fol13].