Some further remarks. A nice aspect of this question is that it provides an example of how some general (and non-constructive) theorems of Functional Analysis often show us the way to a constructive answer to concrete problems. Let's focus on the case of the inclusion $L^\infty(\Omega)$ in $L^1(\Omega)$ (any other non-compactness follows from $L^\infty(\Omega)\to L^p (\Omega)\to L^q(\Omega)\to L^1(\Omega)$ by composition). By the Fréchet-Kolmogorov theorem, a bounded set $B\subset L^1(\Omega)$ (with $\Omega$ a bounded subset of $\mathbb{R}^n$) is relatively compact if and only if it is "equicontinuous-$L^1$," meaning that $$\sup _{f\in B}\ \omega_f(\delta) \ =o(1),\qquad \mathrm {as}\ \delta\to0.$$
Here $\omega_f$ is the "modulus of continuity-$L^1$" of the function $f$, namely $$\omega_f(\delta)= \sup _{|h|\leq \delta}\| f-f(\cdot-h) \|_1\ .$$ This condition is clearly not satisfied by the unit ball $B$ of $L^\infty,$ because for any $\delta$ there is in $B$ a function whose support is disjoint from a $\delta$ translate of it, and in . In fact $\| f-f(\cdot-h) \|_1=|\Omega|$ holds, for instance, for the characteristic function of a suitable measurable set. So this answers the question, and also suggests an answer independent from the FK thm, by exhibiting directly a non-compact sequence of oscillating functions, like in the answer by Denis Serre's answer.
