Another approach uses the representation of the dual $\left[ L^1(\mathbb{R})\right]^\star$ as $L^\infty(\mathbb{R})$ and the Hahn-Banach separation theorem. Namely, to prove that a vector subspace of a Banach space is dense we only need to show that the only continuous linear functional that vanishes on it is the null one.
To do this fix $\phi \in L^\infty(\mathbb{R})$ and suppose that
$$\tag{1}\int_{-\infty}^\infty \phi(x)f(x)\, dx=0, \quad \forall f\in C_c(\mathbb{R}).$$
We claim that $\phi=0$ almost everywhere. Indeed, let $a<b$ be fixed numbers. Approximate the characteristic function $\chi_{[a,b]}$ with a family $\chi^{(\varepsilon)}_{[a, b]}$ of "trapezoidal-like" functions:
We have
$$\int_{-\infty}^\infty \left\lvert \chi_{[a,b]}(x)-\chi_{[a,b]}^{(\varepsilon)}\right\rvert\, dx = 2\varepsilon$$
so
\begin{align}
\left\lvert \int_{-\infty}^\infty \phi(x)\chi_{[a,b]}(x)\, dx - \int_{-\infty}^\infty \phi(x)\chi_{[a, b]}^{(\varepsilon)}(x)\, dx \right\rvert & \le \lVert \phi\rVert_{\infty} \int_{-\infty}^\infty \left\lvert \chi_{[a,b]}(x)-\chi_{[a,b]}^{(\varepsilon)}(x)\right\rvert\, dx \\
&=2\varepsilon \lVert \phi\rVert_\infty.
\end{align}
In particular,
$$\int_a^b \phi(x)\, dx=\lim_{\varepsilon \to 0} \int_{-\infty}^\infty \phi(x)\chi_{[a,b]}^{(\varepsilon)}\, dx,$$
and the last limit is $0$ due to our assumption (1): indeed, every $\chi_{[a, b]}^{(\varepsilon)}$ is a continuous function with compact support. We have thus shown that
$$\tag{2} \int_a^b\phi(x)\, dx=0, \qquad \forall a<b.$$
It is intuitively clear that this can happen only if $\phi=0$ almost everywhere: for a rigorous proof of this you can apply the Lebesgue differentiation theorem or Lemma 1 of this post. This proves the claim.
To conclude we only need to recall that every continuous linear functional $\Lambda \in \left[ L^1(\mathbb{R})\right]^\star$ is of the form
$$\Lambda f= \int_{-\infty}^\infty \phi(x)f(x)\, dx,\qquad f \in L^1(\mathbb{R}),$$
for a unique $\phi\in L^\infty(\mathbb{R})$, and then apply the Hahn-Banach separation theorem. $\square$
A final remark: Even if we did not mention convolutions explicitly, the present proof is not that different in nature from the ones presented above. Both rely on the possibility of approximating "rough" functions (like our $\chi_{[a, b]}$) with "smooth" ones.
Best Answer
Let $B_n$ be the ball with radius $n$, $K_n=C_c^\infty(B_n)$ with its metrizable topology, $\varphi_n\in K_n$ a function with support contained in $B_{n}$ and $\varphi_n(x)=1$ for $x\in B_{n-1}$. First observe that $$ F_n\colon K_n\times K_n \to K_n $$ is a continuous map, which can be easily seen by the defining seminorms for these metric spaces.
Now let $U$ be a convex neighbourhood of $0$, i.e. $U\cap K_n$ is a convex neighbourhood of $0$ in $K_n$ for each $n$. Inductively for each $n$, you can find a $0$-neighbourhood $V_n$ of $K_n$ such that $$ F[V_n,V_n] \subseteq U\cap K_n $$ (by the continuity of $F_n$) and $$ \varphi_k V_n \subseteq V_k\,\,\,\,\, (1\leq k < n).$$ Set $W_n:=V_n\cap K_{n-1}$ and $W$ as the convex hull of $\bigcup_n W_n$. Observe that for each $n$, $W_n$ is neigbourhood of $0$ in $K_{n-1}$, so $W\cap K_{n-1}\supseteq W_n$ is one too, hence $W$ is a neighbourhood of $0$ in $C_c^\infty(\mathbb{R}^d)$. Now $F[W,W]\subseteq U$ would establish the continuity of $F$.
Let $\psi, \chi\in W$, i.e. $\psi=\alpha_1\psi_1+\cdots + \alpha_m\psi_m$ and $\chi=\beta_1 \chi_1 + \cdots + \beta_m \chi_m$ with $\alpha_i, \beta_i\geq 0$, $\sum \alpha_i = \sum \beta_i =1$ and $\psi_i,\chi_i\in V_i$. As $$ F(\psi,\chi)=\psi\cdot \chi = \sum_{i,j} \alpha_i\beta_j \cdot \psi_i\chi_j $$ and $\sum_{i,j} \alpha_i\beta_j = 1$, it it sufficient to verify $\psi_i\chi_j\in U$. Now if $i=j$, $$ \psi_i\chi_i = F(\psi_i,\chi_i)\in F[V_i,V_i]\subseteq U\cap K_i \subseteq U.$$ If $i\neq j$, e.g. $i<j$, then $\psi_i\in V_i$ and $\chi_j\in V_j$ and so $$ \psi_i\chi_j = (\psi_i \varphi_i) \chi_j =\psi_i (\varphi_i\chi_j)\in V_i\cdot V_i \subseteq U\cap K_i \subseteq U.$$