We can ignore the information that $J_\varepsilon$ is a mollifier. All we need is a smooth function with integral one. $J_\varepsilon$ is such a function as proven in a) in the question above.
We will use that $C_c(X)$ is dense in $L^1$ to show that $C_c^\infty(X)$ is also dense in $L^1$ where $X$ is an open subset of $\mathbb{R}$. Let $\epsilon > 0$ and $f \in L^1$. Then by density of $C_c(X)$ there is a $g$ in $C_c(X)$ such that $\| f - g \|_{L^1} < \epsilon$.
Now we need to turn $g$ into a smooth function by convolving it with $J_\varepsilon$. Let $$g_\varepsilon (x) := (J_\varepsilon \ast g ) (x) = \int_\mathbb{R} J_\varepsilon(x - y) g(y) dy$$
Then $g_\varepsilon$ is smooth because $\left ( f \ast g \right )^\prime = f^\prime \ast g = f \ast g^\prime$ and $J_\varepsilon$ is infinitely differentiable.
$g_\varepsilon$ has compact support because if $[-S,S]$ is the support of $g$ and $[-R,R]$ is the support of $J_\varepsilon$ then the support of $J_\varepsilon \ast g$ is contained in $[-S - R, S + R]$ and hence is also compact.
To finish the proof we claim that $\| f - g_\varepsilon \|_{L^1} < \epsilon$:
$$ \| f - g_\varepsilon \| \leq \| f - g \| + \|g - g_\varepsilon \| < \epsilon$$
Where $\| f - g \| < \frac{\epsilon}{2}$ holds because $C_c(X)$ is dense in $L^1$ and $\|g - g_\varepsilon \| < \frac{\epsilon}{2}$ holds because:
$$\begin{align}
\|g - g_\varepsilon \|_{L^1} = \int_X \left |Â g(z) - g_\varepsilon (z)\right | dz
&= \int_X \left |Â g(z) - \int_\mathbb{R} J_\varepsilon(z -y) g(y) dy \right | dz \\
&= \int_X \left |Â g(z)\int_\mathbb{R}J_\varepsilon(y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz\\
&\stackrel{(*)}{=} \int_X \left |Â g(z)\int_\mathbb{R}J_\varepsilon(z - y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz \\
&= \int_X \left |Â \int_\mathbb{R} g(z) J_\varepsilon(z - y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz \\
&\leq \int_X Â \int_\mathbb{R} |Â g(z) J_\varepsilon(z - y) |dy - \int_\mathbb{R} | J_\varepsilon(z -y) g(y) |Â dy dz \\
&= \int_X \int_\mathbb{R} |g(z) - g(y)| J_\varepsilon (z -y) dy dz
\end{align}$$
Where the equality marked with (*) holds because the integral is over all of $\mathbb{R}$ so the shift by the constant $z$ doesn't change the integral and $J_\varepsilon$ is even hence $J_\varepsilon (y) = J_\varepsilon (-y)$.
$g$ is continuous and compactly supported hence it is uniformly continuous and so there exists a $\delta$ such that $|g(z) - g(y)| < \frac{\epsilon}{2 \lambda(X)}$ for all $z,y \in X$ hence by choosing $\varepsilon := \delta$ we get
$$ \int_X \int_\mathbb{R} |g(z) - g(y)| J_\delta (z -y) dy dz < \frac{\epsilon}{2} $$
Note that $\epsilon$ and $\varepsilon$ are not the same.
Best Answer
The argument I know depends on a few facts.
Consider the convolution $$(f*g)(x)=\int_{\mathbb R^d} f(t)\,g(x-t)\, dt.$$
One can show that if $f\in L^p$ and $g\in L^1$, then $$\|f*g\|_p\leq\|f\|_p\|g\|_1.$$ so $f*g\in L^p$.
If $f\in L^p$, $g\in C^r(\mathbb R^d)$ with compact support and $D$ is a mixed partial derivative of order $m$, then $$ D(f*g)(x)=(f*Dg)(x). $$ In particular, $f*g\in C^r(\mathbb R^d)$. So if $g\in C^\infty$, so does $f*g$.
Let $f\in L^p(\mathbb R^d)$, $g\in L^1(\mathbb R^d)$ with $\int_{\mathbb R^d} g=1$. Let $g_n(x)=n^d\,g(nx)$. Then $$\lim_{n\to\infty}\|f-f*g_n\|_p=0.$$
Since the compactly supported functions are dense in $L^p$, we may assume that $f$ is compactly supported. Then the functions $f*g_n$ from above are compactly supported.
Let $f\in L^p(\mathbb R^d)$. Take $g$ to be any compactly supported $C^\infty$ function with $\int_{\mathbb R^d} g=1$. Fix $\varepsilon>0$. There exists $f_0\in L^p(\mathbb R^d)$, with compact support, such that $\|f-f_0\|_p<\varepsilon/2$. By the above steps, we have that $f_0*g_n\in C^\infty_c(\mathbb R^d)$ and $\|f_0-f_0*g_n\|_p<\varepsilon/2$. Then $$\|f-f_0*g_n\|_p\leq\|f-f_0\|_p+\|f_0-f_0*g_n\|_p<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$ Thus we get that $f$ is a limit of compactly supported infinitely differentiable functions.
Finally, it remains to construct a nonzero infinitely differentiable, compactly supported function. Start with $$h_1(t)=\begin{cases} e^{-1/x^2},&\ x>0\\ 0,&\ x\leq0\end{cases}$$ Then $h_1\in C^\infty(\mathbb R)$. Now, given any interval $[a,b]$, form $$h_{a,b}(x)=h_1(x-a)h_1(b-x).$$ Then $h_{a,b}\in C^\infty(\mathbb R)$ with support in $[a,b]$. Now given any box $B=\prod_{j=1}^d[a_j,b_j]\subset \mathbb R^d$, the function $$ g_B(x)=h_{a_1,b_1}(x_1)\cdots h_{a_d,b_d}(x_d) $$ is nonzero, $C^\infty$, and with support in $B$.