Here is a sketch of my proof, but I think many points are still missing. Could you please take a look and complete the proof?
Let $S(R^n)$ be a Schwartz space, which is a Fréchet space. If it has the Heine-Borel property, then by definition it is also a Montel space. Let $F \subseteq S(R^n)$ be a bounded and closed set, and $\{f_n\}$ be a bounded sequence in $F$, then $f_n$ is uniformly bounded in each of its partial derivatives, which gives that $\{f_n\}$ is equicontinuous, and by the mean value theorem all of its partial derivatives are also equicontinuous. Then for each $\alpha$, $\{\partial^\alpha f_n\}$ is a uniformly bounded, equicontinuous set, and thus by the Arzelà-Ascoli theorem, it has a uniformly convergent subsequence, namely, it is sequentially compact. For the Schwartz space $S(R^n)$ which is separable as a metric space, and the functions in it are rapidly decreasing and vanishing at infinity along with all partial derivatives, hence the sequentially compact $F$ is compact.
That is the basic idea but you are ignoring one point. Consider the case $n=1$. Your argument shows that $(f_n)$,$(f_n')$,$(f_n'')$,... all converge uniformly on compact sets but you have to know that the limit of the second sequence is the derivative of the limit of the first, and so on. You need the following:
Lemma: if $f_n \to f$ uniformly and $f_n' \to g$ uniformly then $f$ is differentiable and $f'=g$.
Proof: $f_n(x)=f_n(0)+\int_0^{x} f_n'(t) dt$. This gives
$f(x)=f(0)+\int_0^{x} g(t) dt$. The Lemma follows from this.