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.
Best Answer
We need assumption $\int |f| d\mu < \infty$.
We begin with the following claim:
Proof:
Let $h_n = \min\{g_n, g\}$. We have $h_n \to g$ a.e, $0\leq h_n \leq g$.
By the dominated convergence theorem, we have $\int h_n d\mu \to \int g d\mu$.
In other hand, we have $$h_n = \frac{(g_n + g -|g_n -g|)}{2}\text{ or }|g_n -g| = 2h_n -g_n -g$$ Hence the claim is proved.
By the claim, we have $|f_n| \to |f|$ in $L^1$.
Let $g_n = (|f_n| + f_n)/2$, we have $g_n \geq 0$, $g_n \to (|f| + f)/2$ and $g_n \leq |f_n|$.
By Fatou's lemma, we have $$\liminf \int g_n d\mu \geq \int \frac{|f| +f}2 d\mu.$$ In other hand, $|f_n| -g_n = (|f_n| -f_n) / 2 \geq 0$, hence by Fatou's lemma again, we have $$\liminf \int (|f_n| -g_n) d\mu \geq \int (|f| - \frac{|f| +f}2) d\mu.$$ Since $\int |f_n| d\mu \to \int |f| d\mu$, hence $$\liminf \int (|f_n| -g_n) d\mu = \liminf (\int |f_n| d\mu - \int g_n d\mu) = \liminf \int |f_n| d\mu + \liminf (-\int g_n d\mu) = \int |f| d\mu -\limsup \int g_n d\mu.$$ Hence $$\limsup \int g_n d\mu \leq \int \frac{|f| +f}2 d\mu.$$ Therefore, we have $\int g_n d\mu \to \int \frac{|f| +f}2 d\mu$.
By the claim, we have $g_n \to (|f|+f)/2$ in $L^1$.
Hence $f_n = 2g_n - |f_n| \to 2 \frac{|f| +f}2 -|f| =f$ in $L^1$.