Proposition. Let $f\in C^\infty(\mathbb R^n)$. The following statements are equivalent.
(a) $\displaystyle\|f\|_{\alpha,\beta}:=\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta f(x)|<\infty$ for all $\alpha,\beta\in\mathbb{N}^n$.
(b) $\displaystyle\sup_{x\in\mathbb{R}^n}\|x\|^k|D^\beta f(x)|<\infty$ for all $k\in \mathbb N$ and $\beta\in\mathbb{N}^n$.
(c) $\displaystyle\lim_{|x|\to\infty}\|x\|^kD^\beta f(x)=0$ for all $k\in \mathbb N$ and $\beta\in\mathbb{N}^n$.
(d) $\displaystyle\lim_{|x|\to\infty} x^\alpha D^\beta f(x)=0$ for all $\alpha,\beta\in\mathbb{N}^n$.
As a consequence we have the following
Corollary. Let $f\in C^\infty(\mathbb R^n)$. Then, $f$ is a Schwartz function if and only if $f$ is a rapidly decreasing function.
Proof of the Corollary: To say that $f$ is in the Schwartz space means that $f$ satisfies the condition $(a)$ of the proposition. To say that $f$ is rapidly decreasing means that each $D^\beta f$ tends to $0$ faster than $\|x\|^{-k}$ for all $k\geq 0$ as $|x|\to\infty$ which is precisely the condition $(c)$ of the proposition. $\blacksquare$
Proof of the Proposition:
$\text{(a)}\Rightarrow\text{(b)}$ Let $e_1,...,e_n$ be the canonical basis of $\mathbb{R}^n$. Then
$$\|x\|^k=\left(\sum_{j=1}^n x_j^2 \right)^{k/2}\leq C_{k,n}\sum_{j=1}^n(x_j^2)^{k/2}=C_{k,n}\sum_{j=1}^n|x_j|^k=C_{k,n}\sum_{j=1}^n|x^{ke_j}|$$
for all $x=(x_1,...,x_n)\in\mathbb{R}^n$, where $C_{k,n}$ is a positive constant that depends on $k$ and $n$. Thus,
$$\begin{align}
\sup_{x\in\mathbb R^n}\|x\|^k|D^\beta f(x)|&\leq\sup_{x\in\mathbb R^n}\left(C_{k,n}\sum_{j=1}^n|x^{ke_j}|\right)|D^\beta f(x)|=C_{k,n}\sup_{x\in\mathbb R^n}\sum_{j=1}^n|x^{ke_j}D^\beta f(x)|\\
&\leq C_{k,n}\sum_{j=1}^n\|f\|_{ke_j,\beta}<\infty.
\end{align}$$
$\text{(b)}\Rightarrow\text{(c)}$ There exists a constant $M$ such that $\|x\|^{k+1}|D^{\beta}f(x)|\leq M$ for all $x\in\mathbb R^n$ and thus
$$\left|\|x\|^{k}D^{\beta}f(x)\right|\leq\frac{M}{|x|},\qquad\forall\ x \neq 0.$$
Since the right hand side of the last inequality goes to zero as $|x|\to\infty$, we get the result.
$\text{(c)}\Rightarrow\text{(d)}$ Let $\alpha=(\alpha_1,...,\alpha_n)$ be a multi-index and $k=\alpha_1+\cdots+\alpha_n$. Then
$$|x^\alpha|=\prod_{j=1}^n|x_j^{\alpha_j}|=\prod_{j=1}^n(x_j^{2})^{\alpha_j/2}\leq \prod_{j=1}^n\left(\sum_{r=1}^nx_r^2\right)^{\alpha_j/2}=\|x\|^{k}$$
for all $x=(x_1,...,x_n)\in\mathbb{R}^n$. Thus,
$$|x^\alpha D^\beta f(x)|\leq\left|\|x\|^kD^\beta f(x)\right|,\qquad\forall \ x \in\mathbb R^n.$$
Since the right hand side of the last inequality goes to zero as $|x|\to\infty$, we get the result.
$\text{(d)}\Rightarrow\text{(a)}$ There exists a constant $M$ such that
$$|x^\alpha D^\beta f(x)|\leq 1,\quad\forall\ x\notin \overline{B(0,M)}$$
and thus
$$\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta f(x)|\leq\left\{1,\max_{x\in \overline{B(0,M)}}|x^\alpha D^\beta f(x)|\right\}<\infty.\blacksquare$$
Best Answer
Note that by definition, the derivative of a Schwarz function is a Schwarz function, hence we merely have to prove that each element of the Schwarz space is uniformly continuous.
As suggested in the comments, we can prove that if a function $f\colon\mathbb R^n\to\mathbb R$ is continuous and vanishes at infinity, then $f$ is uniformly continuous. To show this, fix a positive $\varepsilon$. For some $R$, we have $|f(x)| \lt\varepsilon$ provided that $|x|\geqslant R$. The closed ball of $\mathbb R^n$ with radius $2R$ and centered at the origin is compact. Therefore, $f$ is uniformly continuous on this set. There exists a $\delta$ such that if $|x|,|y|\leqslant 2R$ and $|x-y|\lt\delta$, then $|f(x)-f(y)|\lt \varepsilon$. We have to extend this to arbitrary $x$ and $y$ such that $|x-y|\lt\delta$ (that is, not only for those of the ball). We have to treat the cases:
One can actually show that if $f\in\mathcal S(\mathbb R^n)$, then $f$ is Lipschitz-continuous. To see this, fix $x,y\in\mathbb R^n$ and consider the function $g\colon\mathbb R\to\mathbb R$ defined by $g(t):=f(x+ty)$ and use the mean value theorem in order to bound $f(x)-f(y)$.