If $u \in \mathscr{D}^{\prime}(\Omega)$ and $\varphi \in C^{\infty}(\Omega)$, then,
$\operatorname{supp}(\varphi u) \subseteq \operatorname{supp}(\varphi) \cap \operatorname{supp}(u)$
Suppose $x\in \operatorname{supp}(\varphi u)$. Since $x\in \operatorname{supp}(\varphi u)$ is a closed set, then $\operatorname{supp}(\varphi u)$ contains every limit point. Now we must show that $x\in \operatorname{supp}(\varphi)$ and $x\in \operatorname{supp}(u)$. How should I proceed to show that $x\in \operatorname{supp}(\varphi)$ and $x\in \operatorname{supp}(u)$. I was thinking perhaps showing that every convergent sequence in $\operatorname{supp}(\varphi u)$ converges to a limit in $\operatorname{supp}(\varphi u)$, but that will only proof that the set is closed. Any hints, ideas, help would be appreciated.
$\phi u$ is a distribution that acts on test functions $f$ by the rule $$\phi u(f) = u(\phi f).$$
Suppose that $x \notin \mathrm{supp\,} u$. Then there is a neighborhood $V$ of $x$ with the property that $u(f) = 0$ for every test function compactly supported in $V$. If $f$ is such a function so is $\phi f$, so that $\phi u(f) = u(\phi f) = 0$. Thus $x \notin \mathrm{supp\, }\phi u$.
Suppose that $x \notin \mathrm{supp \,}\phi$. Then $x$ has a neighborhood $V$ with the property that $\phi$ vanishes on $V$. If $f$ is a test function compactly supported in $V$ then $\phi f$ vanishes everywhere so that $\phi u(f) = u(\phi f) = 0$. Thus $x \notin \mathrm{supp\, }\phi u$.