Let $f\in\mathcal S'(\mathbb R^n)$ and $\varphi\in\mathcal D(\mathbb R^n)$. Suppose $\varphi(x)=0$ for $x\in U$, where $U$ is a neighborhood of supp$f$. Then, does supp$f\cap$ supp$\varphi=\emptyset$ hold ?
I don't know whether this really hold or not, but I expect this holds.
Now, if I suppose there is $x\in$ supp$f\cap$ supp$\varphi$, I get $\varphi(x)=0$ and $x\in\overline{\{x\mid \varphi(x)\neq 0\}}$. I cannot see a contradiction from here. Could you find ?
The definition of support of $f$ in my textbook is here.
For $f\in\mathcal S'(\mathbb R^n)$ and $G\subset_{\text{open}}\mathbb R^n$, we define $$f=0\ \mathrm{on}\ G\iff \langle f,\varphi\rangle=0\ \mathrm{for\ each}\ \varphi\in\mathcal S(\mathbb R^n)\ \mathrm{with}\ \operatorname{supp} \varphi\subset G.$$ And define supp$f$ as the largest open set on which $f=0.$
And perhaps, the following lemma may help.
Lemma
For $f\in\mathcal S'(\mathbb R^n)$ and $x\in\mathbb R^n$, $x\notin$ supp$f$ if and only if there is an open neighbourhood of $x$, say $U$, such that $\langle f,\varphi\rangle=0$ for all $\varphi\in C_0^\infty(U).$
So if you have that, you have an open set $U$ such that $$\{ \varphi = 0 \} \supset U $$ And $$ supp f \subset U$$ So you have $$(supp \varphi )^c \supset U$$ By passing to complementary $$supp \varphi \subset U^c$$ Hence $$ supp f \cap \sup \varphi \subset U \cap U^c = \varnothing$$
This is an equivalence, for the return just set as a neighborhood $U = (supp \varphi)^c$.