Prove that $\int_E |f_n-f|\to0 \iff \lim\limits_{n\to\infty}\int_E|f_n|=\int_E|f|.$

As Prahlad Vaidyanathan said, the proof is correct. I would probably not use $\varepsilon$ in the first part, instead writing

$$ \left|\int_E|f_n| - \int_E|f| \right| = \int_E \big||f_n|-|f|\big|\le \int_E |f_n-f|\to 0 $$ and invoking the squeeze lemma.

Fatou's Lemma is your friend. By Fatou,

\begin{align*} \int_{E} 2|f| &= \int_{E} \liminf_{n\to\infty} (|f| + |f_n| - |f-f_n|) \\ &\leq \liminf_{n\to\infty} \int_{E} (|f| + |f_n| - |f-f_n|) \\ &= 2\int_{E} |f| - \limsup_{n\to\infty} \int_{E} |f-f_n|. \end{align*}

So it follows that $\limsup_{n\to\infty} \int_{E} |f-f_n| = 0$ and the desired conclusion follows. You may also want to give a look on Scheffé's lemma.