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.
Best Answer
This is no answer to my question, just a slightly weaker version which I was able to prove:
If $f,g\colon X \to \mathbb{R}$ are quasi-integrable with $\int_A fd\mu \leq \int_A g d\mu$ for all $A \in \Omega$, then $f \leq g$ a.e. (so the only difference being that $f$ and $g$ now assume values solely in $\mathbb{R}$).
First we note that it suffices to show $f^+ \leq g^+$ and $f^- \geq g^-$ a.e. Hence without loss of generality we can assume that $f,g \colon X \to [0, \infty)$. For $h \geq 0$ measurable define $\nu_h$ to be the measure with $\nu_h(A)\equiv \int_A h d\mu$. Our condition on $f$ and $g$ then becomes $\nu_f \leq \nu_g$ and we want to prove that this already implies $f\leq g$ a.e. We now note that both $\nu_f$ and $\nu_g$ are sigma-finite: Indeed let $\{A_n\}_n \subset \Omega$ such that $\bigcup A_n = X$ and $\mu(A_n) < \infty$ for all $n$. Define $A_{nm} \equiv A_n \cap \{f\leq m\}$, then $\bigcup_{n,m\geq 1} A_{nm}= X$ and $\nu_f(A_{nm}) \leq m \mu(A_{nm}) <\infty$. This shows sigma-finitess for $\nu_f$ and analogously for $\nu_g$. Now let $\{C_n\}_n$ be a sigma-sequence for both $\nu_f$ and $\nu_g$, meaning $\bigcup C_n= X$ and $\nu_f(C_n), \nu_g(C_n)<\infty$ for all $n$. Define $B_m \equiv \{f>g+1/m\}$ and $C_{nm} \equiv C_n \cap B_m$, then we have $\nu_f(C_{nm})=\int_{C_n} f \chi_{B_m}d\mu \geq \int_{C_n}(g+1/m)\chi_{B_m}\geq \nu_f(C_{nm})+\mu(C_{nm})/m$, implying $\mu(C_{nm})=0$ for all $n,m$, and since $B_m \nearrow B \equiv \{f>g\}$ we have by continuity from below $\mu(C_n\cap B)= lim_{m \to \infty} \mu(C_n\cap B_m)=0$ for all $n$, implying $\mu(B)=0$, which is exactly what we wanted.