I believe I have a proof for the following problem, though I do not rely on the finite measure space criterion. Any advice would be appreciated, especially if there is some connection with this problem to the $L^p$ version of Arzela-Ascoli and the Dunford-Pettis theorem.
Let $\Omega$ be a finite measure space. Prove that if $K \subset L^1(\Omega)$ is compact, then $K$ is equi-integrable.
Fix $\epsilon > 0$. Since $K$ is a compact metric space, it is totally bounded, so there exist $f_1,\dots,f_n$ such that $B(f_1,\epsilon), \dots, B(f_n,\epsilon/3)$ cover $K$. From standard measure theory, any given $f \in K$ is equi-integrable on its own (i.e., absolute continuity of the Lebesgue integral), so any finite set of functions is equi-integrable. Thus there exists $\delta > 0$ such that $h < \delta$ implies $\lVert{\tau_h f_i - f_i\rVert}_1 < \epsilon/3$ for each $i = 1,\dots,n$.
We claim that this choice of $\delta$ works. Indeed, for any $f \in K$, there exists $N$ such that $\lVert{f - f_N\rVert}_1 < \epsilon/3$. Then for $h < \delta$, $$ \lVert{\tau_h f - f\rVert} \leq \lVert{\tau_h f - \tau_h f_1\rVert} + \lVert{\tau_h f_1 - f_1\rVert} + \lVert{f_1- f\rVert} = 2\lVert{f - f_1\rVert} + \lVert{\tau_h f_1 - f_1\rVert} < \epsilon, $$ as desired.
EDIT: $\tau_h(f) = f(x+h)$. I also realized my definition of equi-integrability is a bit different: for $\epsilon > 0$ there exists $\delta $ such that $\int_A |f| < \epsilon$ whenever $|A|<\delta$ for all $f \in K$. Does what I prove suffice though?
I don't see how the shift operators come into play. You don't need them.
As you said, there exist $f_1,\ldots,f_n$ such that the open balls $B(f_j,\epsilon/2)$, $j=1,\ldots,n$, cover the compact set $K$. Since the finite set $\{f_1,\ldots,f_n\}$ is uniformly integrable, we can choose for given $\epsilon>0$ a constant $\delta>0$ such that
$$\int_{A} |f_j| < \frac{\epsilon}{2}$$
for all measurable sets $A$ with $|A|<\delta$ and $j=1,\ldots,n$. Now if $f \in K$, then we can find $j \in \{1,\ldots,n\}$ such that $\int |f-f_j| < \epsilon/2$. Hence, by the triangle inequality
$$\int_A |f| \leq \int_A |f-f_j| + \int_A |f_j| \leq \int |f-f_j| + \int_A |f_j| < \epsilon$$
for any measurable set $A$ with $|A|<\delta$. This proves that $K$ is equi-integrable.