I'll assume we are talking about functions on the real line. Fix $f\in C_0(\mathbb R)$. The fact that $f$ is $C_0$ allows us to write it as a uniform limit of continuous functions with compact support$^1$. So now we can assume without loss of generality that $f$ is continuous with compact support.
Start with $h_0(x)=\begin{cases}e^{-1/x^2},&\ x>0\\0,&\ x\leq0\end{cases}$.
Next you notice that $h(x)=h_0(x)h_0(1-x)\in C^\infty_K$, with support in $[0,1]$. We can normalize it so that $\int_{\mathbb R} h=1$. Then, for each $\varepsilon>0$, you take
$$
h_\varepsilon(x)=\frac1\varepsilon\,h\left(\frac x\varepsilon\right)
$$
and you note that $\int_{\mathbb R}h_\varepsilon=\int_{\mathbb R}h=1$.
Now form the convolutions
$$
f_\varepsilon(x)=\int_{\mathbb R}f(t)\,h_\varepsilon(x-t)\,dt.
$$
It is not hard to show that because $h_\varepsilon\in C^\infty$ we have $f_\varepsilon\in C^\infty$; and because $f$ and $h_\varepsilon$ have compact support, so does $f_\varepsilon$.
Finally, given $\varepsilon>0$, as $f$ is continuous with compact support, it is uniformly continuous. So given $c>0$ there exists $\delta>0$ such that $|f(x)-f(y)|<c$ if $|x-y|<\delta$. Then
\begin{align*}
|f_\varepsilon(x)-f(x)|&=\left|\int_{\mathbb R} [f(x-t)-f(x)]\,h_\varepsilon(t)\,dt\right|
\leq\int_{\mathbb R} |f(x-t)-f(x)|\,h_\varepsilon(t)\,dt\\
&\leq c\,\int_{|t|<\delta} h_\varepsilon(t)\,dt
+2\|f\|_\infty\,\int_{|t|\geq\delta}h_\varepsilon(t)\,dt\\
&=c
+2\|f\|_\infty\,\int_{|t|\geq\delta}h_\varepsilon(t)\,dt.
\end{align*}
with $\delta$ fixed, the last integral goes to zero as $\varepsilon\to0$ (because the support of $h_\varepsilon$ is contained in $[0,\varepsilon]$). Thus
$$
\limsup_{\varepsilon\to0}|f_\varepsilon(x)-f(x)|\leq c
$$
for all $x$ and all $c>0$. So $\|f_\varepsilon-f\|_\infty\to0$.
- Fix $f\in C_0(\mathbb R)$. For each $n\in\mathbb N$ there exists $N>0$ with $|f(x)|<\tfrac1n$ when $|x|>N$. Let $g_n$ be continuous, with $0≤g_n≤1$, $g_n(x)=1$ when $|x|≤N$ and $g_n=0$ when $|x|>N+1$. Then $fg_n\in C_c(\mathbb R)$ and $$|f-fg_n|=|f(1-g_n)|<\tfrac1n,$$ so $fg_n\to f$ uniformly.
Best Answer
Let us write that norm out, actually omitting the $p$-th root because it's irrelevant for "tends to 0":
$$\|f(x+h)-f(x)\|_p^p=\int_{\mathbb{R}^d}|f(x+h)-f(x)|^pdx.$$
First assume $f\in\mathcal{C}_c(\mathbb{R}^d)$. That integral then is only on the union of $\mathrm{supp}(f)$ and $\mathrm{supp}(f)-h$, for outside that the integrand is zero. If we assume $|h|<\delta$ for some $\delta$, we can enlarge that to an integral over $X_\delta:=\{x:d(x,\mathrm{supp}(f))<\delta$, for both pieces of the union are included in that. This is a compact set, and now it only depends on $\delta$, not directly on $h$, so we can try dominated convergence. By continuity, for all $\varepsilon$ we can find $\delta$ such that $|h|<\delta\implies|f(x+h)-f(x)|<\varepsilon$. So if $|h|<\delta$, the integrand is always less than $\varepsilon^p$, and the integral will be less than $\epsilon^p\lg(X_\delta)$, so:
$$\|f(x+h)-f(x)\|_p\leq\varepsilon\sqrt[p]{\lg(X_\delta)}.$$
For $\delta\to0$, the RHS goes to zero because $X_\delta$ tends to be a single point, so for $h\to0$ we can bound the LHS from above with the RHS that goes to zero, hence the LHS goes to zero, as we wanted. If $f$ is not continuous with compact support, by the density theorem we find $g_\epsilon$ such that $\|f-g_\epsilon\|_p<\epsilon$, for any $\epsilon$, and then:
$$\|f(x+h)-f(x)\|_p\leq\|f(x+h)-g_\epsilon(x+h)\|_p+\|g_\epsilon(x+h)-g_\epsilon(x)\|_p+\|g_\epsilon(x)-f(x)\|_p,$$
but the middle term tends to zero and the other two are both equal to $\|f-g_\epsilon\|$, so basically we have bounded our norm from above with any $3\epsilon$, any time $h$ is chosen such that $\|g_\epsilon(x+h)-g_\epsilon(x)\|_p\leq\epsilon$. Hence, the norm goes to zero, completing the proof.