If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$.
Take $\phi\in \mathcal C^\infty_c(\Bbb R)$, such that $\phi\ge 0$ and $\int_{\Bbb R}\phi(x)dx=1$. Let's call $\phi_n(x) = n\phi(nx)$.
On the one hand, this is a regularising sequence. Therefore, $\phi_n\ast f\to f$ in the sense of $L^1$ (the proof of this fact, however, relies on density of $\mathcal C _c$ in $L^1$.)
On the other hand, since $\phi_n(y-x)\in \mathcal C^\infty_c(\Bbb R)$ for a fixed $y$, we can write by our hypothesis $$0=\int_{\Bbb R}\phi_n(y-x)f(x)dx.$$
Thus, $f=0$ in the sense $L^1$.