Here are two alternative methods:
- You have an explicit formula for $u * \phi$ : it is given by $u* \phi(x) = \langle u,\phi(x-y)\rangle_y$, and you can "differentiate under the bracket". Now if $(\phi_n)_n$ goes to $0$ in $C^\infty_c(\mathbb{R}^d)$, you have for any $\alpha\in\mathbb{N}^d$ that $\|u* \partial^\alpha \phi_n\|_\infty = \|\langle u,\partial^\alpha \phi_n(\cdot-y)\rangle_y\|_\infty$ which converges to $0$ (you can one work on a fixed compact $K$ because the sequence converges in $C^\infty_c(\mathbb{R}^d)$ and $u$ is assumed compactly supported).
- You can first try to prove the (weaker) fact : $(u* \phi_n)_n$ is bounded in $C^\infty_c(\mathbb{R}^d)$ whenever $(\phi_n)_n$ is. Once you know that the conclusion follows thanks to the Montel property satisfied in $C^\infty_c(\mathbb{R}^d)$ : bounded sets are relatively compact (this is simply an iterate use of Ascoli's Theorem). Since $(u * \phi_n)_n$ obviously converges to $u*\phi$ for weaker topologies (for instance the one of $C^\infty_c(\mathbb{R}^d)$), the latter is the only possible limit point: the whole sequence converges to it.