Let $F \in D'(\mathbb{R})$. If there exists $f \in D(\mathbb{R})$ which vanishes on $supp(F)$, but $F(f)\neq0$, then how can one prove that $supp(F)$ contains an isolated point?
For example, if $F=\delta'$, and $f(x)=x$, than $F(f)=-1$, $supp(F)=\{0\}$ and $f$ vanishes on it.
If $supp(F)=[0,1]$ or even the Cantor set (!), then such function $f$ does not exist.
If $f(x_n)=0$ for a monotone sequence $x_n\to x_\infty$ the mean value theorem gives you $y_n$ between $x_n$ and $x_{n+1}$ such that $f'(y_n)=0$. The continuity of $f'$ then implies $f'(x_\infty)=0$. Repeating this argument implies $f^{(n)}(x_\infty)=0$ for all $n$. If $supp(F)$ has no isolated points, all derivatives vanish there for $f\in D(\mathbb R)$ which vanishes on $supp(F)$. This is enough to conclude $F(f)=0$ (as can be seen in most text books about distribution theory).