The measurable simple functions $\mathcal{S}$ are dense in $L^p(X,\mathcal{A,\mu}).$ The key result on which this is based is the following theorem.
Theorem. Let $(X,\mathcal{A})$ a measurable space, $f\colon X\to \mathbb{\overline{R}}$. Then exists a sequence $\{s_n\}$ of simple functions on $X$ such that $\lim_{n\to\infty} s_n(x)=f(x)$ in $X$. Moreover,
$(i)\;$ if $f$ is measurable, then $s_n$ is measurable for all $n$;
$(ii)\;$ if $f$ is nonegative, the sequence $\{s_n\}$ is increasing and we have $$0\le s_n\le f\quad n\in\mathbb{N}.$$
$(iii)\;$ if $f$ is bounded, then the convergence is uniform.
Now, in the proof it is supposed that, at first, that $f$ is bounded and nonnegative; then it is built $$s_n:=\sum_{k=0}^{2^n-1}\frac{k}{2^n}\chi_{E_k^n},$$ where $$E_k^n:=\bigg\{x\in X\;\bigg|\; \frac{k}{2^n}\le f(x)<\frac{k+1}{2^n}\bigg\}.$$
If $f$ it is not bounded(non negative), then it is built $$s_n=n\chi_{E_I^n}+\sum_{k=0}^{n2^n-1}\frac{k}{2^n}\chi_{E_k^n},$$ where $$E_I^n:=\{x\in X\;|\; f(x)\ge n\}.$$
Question. In light of the above, since the $s_n$ coefficients are always rational, can I conclude in the same way that the simple functions with rational coefficients (obviously with limited support) are dense in $L^p(X)$? If I can't finish this, how do I show it then?
Yes, the theorem you cite leads to the result on density. Suppose $1\le p <\infty$ and $f\in L^p.$ Write $f= f^+-f^-.$ Then there exist simple nonnegative $s_n,t_n$ with rational coefficients such that $0\le s_n \le f^+,$ $0\le t_n \le f^-$ with $ s_n \to f^+,$ $ t_n \to f^-$ pointwise everywhere. Note that
$$|s_n-f^+|^p = (f^+ -s_n)^p \le (f^+)^p\,\,\text{everywhere}.$$
Thus by the DCT, $\|s_n-f^+\|_p \to 0,$ i.e., $s_n\to f^+$ in $L^p.$ Similarly, $t_n\to f^-$ in $L^p.$ It follows that $s_n-t_n \to f^+ - f^-=f$ in $L^p.$ Since each $ s_n+t_n$ is a simple function with rational coefficients, we're done.
I'll leave the $p=\infty$ case to you for now. Ask if you have any questions.