Show that if $F$ is a distribution, such that $|\xi|^{2}F=0$, then support (F) is subset of $\{0\}$.
My approach: Recall that, the support of a distribution $F$ is the complement of the union of all open sets $U\in \mathbb{R}^n$ such that $F(\varphi) = 0$ (for $\varphi\in \mathcal{D}_{K}$ with compact $K\subset U$).
So, if $F$ is a distribution such that $|\xi|^{2}F=0$, then $$|\xi|^{2}F(\varphi)=\int_{\mathbb{R}^{n}}{ |\xi|^{2}f(x)\varphi(x)dx }=0$$
Wlog, we assume that $F\neq 0$ and the support of $\varphi$ is not in $\{0\}$. Then, the unique way that the above integral will be $0$, is that in $\xi=0$ we have $f(\xi)=0$.
In your proof you are assuming that $F$ is a regular distribution, i.e. given by a function. This is not necessarily true.
Hint: If $\phi \varphi\in \mathcal{D}_{K}$ is such that $\operatorname{supp}(f) \subset \mathbb R^d \backslash \{ 0\}$ then show that there exists some $\psi \in \mathcal{D}$ such that $$\varphi=|\xi|^{2} \psi$$
Then $$F(\varphi)=|\xi|^2F(\psi)$$