The proof is incorrect. In the fist part of the proof, the $\delta$ you picked does not guarantee that for all measurable sets $A$, $\mu(A) < \delta$ implies $|\int_A f_n\, d\mu| < \epsilon$. The second part of the proof is almost correct, but the condition $\lim\limits_{M\to \infty} \mu(E_M) = 0$ follows from the assumption $\sup_n \int |f_n|\, d\mu < \infty$, not that $\mu$ is finite. Indeed, if $\alpha := \sup_n \int |f_n|\, d\mu < \infty$, then $\mu(E_M) \le \frac{\alpha}{M}$ for all $M > 0$; as a consequence, $\lim\limits_{M\to \infty} \mu(E_M) = 0$. If you make this fix, then the second part of the proof will be correct.
To prove the forward direction, fix $\epsilon > 0$ and choose $M > 0$ such that
$$\sup_n \int_{E_M} |f_n| \, d\mu < \frac{\epsilon}{2}.$$
Then
$$\int |f_n|\, d\mu = \int_{E_M} |f_n|\, d\mu + \int_{E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(X) < \infty$$
for all $n\in N$.
Set $\delta = \frac{\epsilon}{2M}$. For all measurable sets $A$, $\mu(A) < \delta$ implies
$$\left|\int_A f_n\, d\mu\right| \le \int_A |f_n|\, d\mu = \int_{A\cap E_M} |f_n|\, d\mu + \int_{A\cap E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(A) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Since $\epsilon$ was arbitrary, $\{f_n\}$ is uniformly absolutely continuous.
Check $E = (0, 1]$, $f_n(x) = \frac{n}{\log n} I_{(0, n^{-1})}(x), n = 1, 2, \ldots$. And $m$ is the Lebesgue measure on $E$.
Best Answer
If $\mu$ is counting measure on a finite set then uniform absolute continuity trivially holds for any sequence $(f_n)$: take $\delta <1$. Obviously, $(f_n)$ need not be $L^{1}$ bounded.