Let $\mathcal C_0 (\mathbb R^d)$ be the space of real-valued continuous functions on $\mathbb R^d$ that vanish at infinity. Let $\mathcal C_c^\infty (\mathbb R^d)$ be the space of real-valued smooth functions on $\mathbb R^d$ with compact supports. By Stone–Weierstrass theorem, $\mathcal C_c^\infty (\mathbb R^d)$ is dense in $\mathcal C_0 (\mathbb R^d)$ w.r.t. the topology of $\| \cdot \|_\infty$.
Fix $f \in \mathcal C_0 (\mathbb R^d)$ and $\varepsilon>0$. Is there an explicit construction of $g \in \mathcal C_c^\infty (\mathbb R^d)$ such that $\|f-g\|_\infty <\varepsilon$?
Thank you so much for your elaboration!
(1) Let $\psi(x)$ be a smooth function satisfying $0\leq \psi(x)\leq 1$ and $$\begin{align} \psi(x)&=1,\quad-1\leq x\leq 1,\\ \psi(x)&=0,\quad |x|\geq 2. \end{align}$$ Denote $\psi_\delta=\psi(\frac{x}{\delta})$. For any $f\in\mathcal C_0(\mathbb R)$, $$f\cdot \psi_\delta\in \mathcal C_c(\mathbb R),\quad \Vert f-f\cdot \psi_\delta\Vert_\infty\leq\sup_{|X|\geq \delta} |f(x)|\to 0, \quad \delta\to \infty.$$
(2) Let $\phi(x)$ be a smooth function with support in $(-1,1)$ satisfying $$\begin{align} \int \phi=1. \end{align}$$ Then $\phi_\varepsilon(x)=\frac{1}{\varepsilon}\phi(\frac{x}{\varepsilon})$ is a family of good kernels. For $g\in C_c(\mathbb R)$, we have $$g*\phi_\varepsilon\in \mathcal C_c^\infty(\mathbb R),\quad \Vert g-g*\phi_\varepsilon\Vert_\infty\to 0,\quad \epsilon\to 0.$$
Combine (1) and (2) together, you will get the function required. The general case for $\mathbb R^d$ is similar.