I'm aware of Classifying the compact subsets of $L^p$ and have seen similar posts, but I haven't seen an example of a compact set on an $L^p$ space. Trying to think of a possible simple example, and as it is used as a counterexample of compacity in the space of continuous functions, I asked myself if the set $S=\lbrace f:[0,1]\rightarrow [0,1] : ||f||_p<\infty \rbrace$ is compact on $L^p([0,1])$.
Applying the Frechet-Kolmogorov theorem feels too general to apply and makes me think that it may not be compact, but can't come up with a counterexample apart from the ones used for the set of continuous functions on $[0,1]$. Do you have any hint?
Let $$f_n(x)={1\over 2}[1+\cos(n\pi x)]$$ Then $f_n$ satisfy the requirements although the sequence does not admit an accumulation point. It suffices to show that $g_n(x)=\cos(n\pi x)$ do not accumulate. We have $$\int\limits_0^1\cos(\pi nx)\cos(k\pi x)\,dx=0,\ n>k$$ Any accumulation point $g$ would satisfy $$\int\limits_0^1g(x)\cos(k\pi x)\,dx =0,\ k\ge 0$$ Thus $g=0.$ However $$\int\limits_0^1|\cos(n\pi x)|^p\,dx ={1\over n\pi}\int\limits_0^{n\pi}|\cos x|^p\,dx={1\over \pi}\int\limits_0^{\pi}|\cos x|^p\,dx$$ therefore $g$ cannot be equal $0,$ a contradiction.