I'm not sure you're stating all conditions of Lusin's theorem correctly - the one I'm looking at (from Real Analysis by Gerald Folland) assumes that $\mu$ is a Radon measure on a locally compact Hausdorff space $X$, and concludes that there exists a compactly supported function $g$ that coincides with the given $\varphi$ on a set of measure $< \varepsilon$. If $\varphi$ is bounded, then $g$ can be chosen so that $\| g \|_\infty \leq \| \varphi \|_\infty$.
Anyways, here are some hints: it suffices to show that any characteristic function $\chi_E$ (where $E \subseteq X$) can be approximated arbitrarily well in $L^p$-norm by compactly supported functions (since simple functions are dense in $L^p$ for $1 \leq p < \infty$). Showing this is a simple matter of estimating
$$ \int_X |\chi_E(x) - \varphi(x)|^p \,\mathrm{d}\mu(x)$$
Note that, since $\chi_E$ is bounded, we can choose $g$ such that $\| g \|_\infty \leq \|\chi_E \|_\infty = 1$, and hence $\| \chi_E - \varphi \|_\infty \leq 2$.
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 $\epsilon>0$
Let $g \in C_c$ such that $(\int |g-f|^p)^{\frac{1}{p}} <\frac{\epsilon}{3}$
Exists $\delta >0$ such that $(\int|g_x-g_y|^p)^{\frac{1}{p}}<\frac{\epsilon}{3}\forall x,y :|x-y|<\delta$
By Minkowski's inequality we have that and some changes of variable,we have that:
$\forall x,y:|x-y|<\delta$ $$(\int |f_y-f_x|^p)^{\frac{1}{p}} \leq (\int|f_y-g_y|)^{\frac{1}{p}} (\int |g_y-g_x|^p)^{\frac{1}{p}}+(\int|g_x-g_x|^p)^{\frac{1}{p}}<\epsilon$$