Let $\varphi\in L^1(\Bbb R^n)$ and positive, such that $\int_{\Bbb R^n}\varphi=1$.
Then set $$\varphi_t(x):=\frac1{t^n}\varphi(x/t)$$ for $t>0$; in such a way, $\varphi_t$ has the same properties of $\varphi$.
I have to prove that $$ \varphi_t* f(x)\to f(x)\;\;\;\mbox{as}\;\;t\to0 $$ for every $f\in L^p(\Bbb R^n)$, with $1\le p<\infty$.
There was a hint, which suggested to approximate $f$ by taking a $g$ continous with compact support such that $||f-g||_p<\varepsilon$ and next writing \begin{align*} |\varphi_t* f(x)-f(x)| &\le |\varphi_t* (f-g)(x)|+|\varphi_t* g(x)-g(x)|+|g(x)-f(x)| \end{align*} Now we know that second summand goes to $0$ as $t\to0$, by the regularity and boundedness of $g$.
For the third summand: since we are checking pointwise convergence, once we fixed $x\in\Bbb R^n$, we can take $g$ such that $g(x)=f(x)$, so the third summand vanish: is this correct?
The problem comes from the first summand; it is controlled by $$ M(f-g)(x):=\sup_{r>0}\frac1{|B(x,r[|}\int_{B(x,r[}|f(y)-g(y)|\,dy $$ this is not trivial: let's accept this. We know that $M$ is an operator bounded on $L^p$, but how can we deduce from this a bound which is small as we want (exploiting the $\varepsilon$ distance in the $p$-norm)?
You actually want to obtain $L^p$ convergence, not pointwise which wouldn't make sense for an $L^p$ function anyway. Assuming the result for $f\in C^0$ as you have and using the hint to choose $g∈ C^0$ with $‖f-g‖_{L^p} < ε$, $$ ‖\phi_t * f - f‖_{L^p} \leq ‖\phi_t *(f - g)‖_{L^p} + ‖ \phi_t * g - g ‖_{L^p} + ‖f-g‖_{L^p} $$ The second and third terms are OK; the second because I assumed you can do it, and the third is by the choice of $g$. $M$ being a bounded operator $L^p → L^p$ means that for some $C>0$, $$ ‖Mf‖_{L^p} \leq C‖f‖_{L^p}$$ Just apply this to $M(f-g)$ to get (by our assumption $f-g∈ L^p$) $$ ‖M(f-g)‖_{L^p} \leq C‖f-g‖_{L^p} \leq Cε $$ Which is enough because we have the a.e. control $|\phi_t * (f-g)| \leq M(f-g)$.
As @EricThoma said in the comments, if you know the Young's inequality for convolutions $\|f*g‖_{L^p} \leq ‖f‖_{L^1} ‖g‖_{L^p}$ then you don't ned to worry about maximal functions. There's a proof here.