Edit: I missed some things in my first answer and updated my answer accordingly.
Edit 2: Not only that I missed some things in my first answer, I missed the point completely. I leave the answer here for instructional purposes. Tasadduk's answer is definitely the way to go about 2).
I agree with mac that the first solution is almost ok, but some extra care has to be taken (the supremum might be infinite and even if one could argue that $\infty \cdot 0 = 0$ in measure theory, a clear cut argument is certainly preferable to a convention).
I would first prove that for a non-negative and bounded functions $f$ we have $\int_{A} f \,d\mu = 0$, for this you can use your argument. The monotone convergence theorem then implies that it holds for all non-negative measurable functions, simply by approximating $f$ by $f_{n} = \min{\{f,n\}}$ from below.
Remark. The equality $\int_{A} f = 0$ holds true for all measurable $f$, because a general $f$ can be written as $f = f_+ - f_-$ with $f_{+}(x) = \max{\{f(x),0\}}$ and $f_{-}(x) = \max{\{-f(x),0\}}$ and use the definition $\int_{A} f\,d\mu = \int_{A} f_+ \,d\mu - \int_{A} f_-\,d\mu$ provided at least one of the integrals of the right hand side is finite.
As mac also pointed out, there are quite a few problems in the second proof and I don't see how to save it using your idea.
Here's how I would do it. Let $A_{t} = \{|f| \gt t\}$ and note that for $s \lt t$ we have $A_{s} \supset A_t$. Define
$$a_{t} = \int_{\{|f| \gt t\}} |f|\,d\mu = \int_{A_t} |f|\,d\mu.$$
Since $|f| \geq 0$ and $A_{s} \supset A_{t}$ we have $a_{s} \geq a_{t}$ for $s \leq t$. We want to show that $a_{t} \to 0$ as $t \to \infty$. By monotonicity of $t \mapsto a_{t}$, it suffices to show that $a_{n} \to 0$ as $n \to \infty$ runs through the natural numbers. Since $a_{n} \geq 0$ and $a_{n}$ is monotonically decreasing, the limit $a = \lim_{n \to \infty} a_{n}$ exists, and we want to show that $$ a = 0.$$
Now let $f_{n}(x) = \min{\{|f(x)|,n\}}$ and note that $f_{n} \to |f|$ pointwise (and monotonically). By the monotone convergence theorem (or the dominated convergence theorem, applicable since $f_{n} \leq |f|$ and $|f|$ is integrable, if you prefer) we have
$$ \int |f|\,d\mu = \lim_{n \to \infty} \int f_{n}\,d\mu.$$
On the other hand $n \leq |f|$ on $A_{n}$ and $0 \leq f_{n} \leq |f|$ on $\Omega$ imply
$$ \int f_{n}\,d\mu = \int_{A_{n}} n \,d\mu + \int_{\Omega \smallsetminus A_{n}} f_{n}\,d\mu \leq \int_{A_{n}} |f|\, d\mu + \int |f| \,d\mu = a_{n} + \int |f|\,d\mu.$$
Passing to the limit on both sides of the estimate $\int f_{n} \leq a_{n} + \int |f|$ we get
$$\int |f|\,d\mu = a + \int |f|\,d\mu$$
and as $\int |f|\,d\mu \lt \infty$ we conclude $a = 0$, as we wanted.
This can be proven without reference to Fatou's Lemma or integration.
You had the right idea. Let $F_n = \bigcap_{k=n}^\infty E_k$, and note that $F_n \subseteq F_{n+1}$ for all $n \geq 1$. Because of this nesting, it follows** that $$\mu(\liminf E_n) = \mu\left(\bigcup_{n=1}^\infty F_n \right) = \lim_{n\to\infty} \mu(F_n).$$
On the other hand, as you noted, $F_n \subseteq E_k$ for every $k \geq n$, and so $$\mu(F_n) \leq \inf_{k \geq n}\ \mu(E_k).$$
Now take limits of both sides.
** Here, I am using the fact (which you are encouraged to prove yourself) that if $E_n \subset E_{n+1}$ for every $n \geq 1$, then $\mu(\bigcup_{n=1}^\infty E_n) = \lim \mu(E_n)$.
Best Answer
Let $f_k = f\chi_{E_k}$. Note $f_k \to 0$ almost everywhere since $\mu(E_k) \to 0$. Also, $|f_k| \le |f|$ which is integrable, so Dominated Convergence Theorem gives you the result.