Yes, the convolution of an integrable function $f$ with compact support, and a Schwartz class function $g$ belongs to the Schwartz space again.
Since all derivatives of Schwartz class functions belong to the Schwartz space, in particular are bounded, the convolution
$$(f\ast g)(x) = \int f(y)g(x-y)\,dy$$
is smooth, since the dominated convergence theorem allows differentiating under the integral arbitrarily often. (Since the difference quotients of $\partial^\alpha g$ converge uniformly on $\mathbb{R}$, and the support of $f$ is compact, one can get that result also without the dominated convergence theorem.)
So only the decay remains to be checked. Choose $K > 0$ such that $\operatorname{supp} f \subset [-K,K]$. Since $g\in \mathcal{S}(\mathbb{R})$, for every $\alpha,m\in\mathbb{N}$ there is a constant $C_{\alpha,m}$ such that
$$\lvert \partial^\alpha g(x)\rvert \leqslant \frac{C_{\alpha,m}}{(1+\lvert x\rvert)^m}$$
for all $x\in\mathbb{R}$.
Then for $\lvert x\rvert \geqslant 2K$ we have
$$\begin{align}
\lvert \partial^\alpha(f\ast g)(x)\rvert &= \left\lvert \int_{-K}^K f(y) \partial^\alpha g(x-y)\,dy \right\rvert\\
&\leqslant \int_{-K}^K \lvert f(y)\rvert\, \lvert \partial^\alpha g(x-y)\rvert\,dy\\
&\leqslant \int_{-K}^K \lvert f(y)\rvert \frac{C_{\alpha,m}}{(1+\lvert x-y\rvert)^m}\,dy\\
&\leqslant \int_{-K}^K \lvert f(y)\rvert \frac{C_{\alpha,m}}{\left(1 + \frac{\lvert x\rvert}{2}\right)^m}\,dy\\
&= \frac{2^mC_{\alpha,m}}{(2+\lvert x\rvert)^m}\int_{-K}^K\lvert f(y)\rvert\,dy\\
&\leqslant \frac{C'_{\alpha,m}}{(1+\lvert x\rvert)^m},
\end{align}$$
where $C'_{\alpha,m} = 2^m\lVert f\rVert_{L^1}C_{\alpha,m}$.
Since $(1+\lvert x\rvert)^m \partial^\alpha(f\ast g)(x)$ is continuous, it is bounded on the compact set $[-K,K]$, hence we have
$$(1+\lvert x\rvert)^m\lvert\partial^\alpha(f\ast g)(x)\rvert \leqslant \tilde{C}_{\alpha,m}$$
for all $x\in\mathbb{R}$ and some constant $\tilde{C}_{\alpha,m}$.
So $f\ast g$ is a smooth function such that $x^m\partial^\alpha(f\ast g)(x)$ is bounded for all $\alpha,m\in\mathbb{N}$, and that means precisely $f\ast g\in \mathcal{S}(\mathbb{R})$.
The generalisation to $\mathbb{R}^n$ is immediate.
I assume $f$ is supposed to be compactly supported.
Start with a smooth function $g_1$ so that $|f-g_1| < \epsilon$ everywhere. Let $g_2 = g-2\epsilon$, so that $\epsilon < f-g_2 < 3 \epsilon$ everywhere. Now let $\psi : \mathbb{R} \to \mathbb{R}^{\ge 0}$ be a smooth approximation of the "positive part" function, such that $\psi(x)=0$ for all $x \le 0$, and $|\psi(x) - x| < \epsilon$ for all $x \ge 0$. Set $g_3 = \psi \circ g_2$ and note that $g_3 \ge 0$ everywhere.
Now consider two cases:
if $g_2(x) \ge 0$, then $|g_3(x) - g_2(x)| < \epsilon$, so we have $0 < f(x)-g_3(x) < 4\epsilon$.
if $g_2(x) < 0$, so that $g_3(x) = 0 > g_2(x)$, then $f(x) - g_3(x) < f(x) - g_2(x) < 4 \epsilon$, and since $f(x) \ge 0$, we still have $f(x) - g_3(x) \ge 0$.
If you want, go back and replace all $\epsilon$ by $\epsilon/4$.
Note that since $f$ is compactly supported, the inequality $0 \le g(x) \le f(x)$ implies that $g_3$ is too.
Best Answer
Based on the comments above, it seems you're misparsing "smooth function with compact support" as "function with compact support which is smooth on its support." This is not what it means: the requirements for $f:\mathbb{R}^n\rightarrow\mathbb{R}$ to be in $C_c^\infty(\mathbb{R}^n)$ are precisely that
$f$ be smooth in the usual $\mathbb{R}^n$-sense, and
$cl(\{x:f(x)\not=0\})$ be compact (also in the usual $\mathbb{R}^n$-sense).
In particular, the support of such an $f$ can never be discrete since for any smooth (or even continuous) $f:\mathbb{R}^n\rightarrow\mathbb{R}$ the set $\{x: f(x)\not=0\}$ will be open. In order to be the support of a continuous (let alone smooth) function, a compact set $K$ must satisfy $$K=cl(int(K)).$$ It's a good exercise to check whether this necessary condition is also sufficient.