You can write $dV_g' = f dV_g$ where $f\to 1$ in $C^1$. Then $div_g(X)(f-1) dV_g = div_G(X(f-1))dV_g - gX,\nabla (f-1))dV_g$. The first term integrated on $\Omega$ is $o(1)$ since $|X(f-1)| \to 0$ in $C^0$ and then the second is also $o(1)$ since $f\to 1$ in $C^1$ and $|X|\leq 1$. The issue raised by @OrangeMushroom can be fixed by shrinking $X$ a bit to compensate for the fact that $g'$ is measuring things slightly bigger/smaller than $g$ (but this stretch factor is $1+o(1)$ so goes away.

Yes, you should be able to construct such a metric. Form "bubbles" on $R^2$ i.e. spheres connected sum with the flat plane to produce a small neck. It's clear that the neck is very good isoperimetrically. Now, make two spheres that contain volume V together, and have total neck area A, and make another single sphere with volume V and area A. You should be able to argue that the isoperimtric problem at this volume V has a solution given by two disks and also one given by one. It may take a bit of work to make this rigorous, but I am pretty sure it can be done by a limit argument..

p. 510 of Lax's paper seems to give Szegő's proof in full detail

@TarekAcila, You need a uniform $M$ that works for all $(x,\sigma)$.

Nice to hear the notes were useful. I wrote a sketch of the proof below, please feel free to let me know if you want more details anywhere.

Yes this is still true. Are you asking for the proof or a reference?

That's what I meant yes!

The positive part of $f'(u)$ can be handled with Fatou's lemma no matter how $f'(u)^+$ behaves.

I think you're missing a diagonal argument.

Isn't $C^0$ sufficient? The equation you write doesn't depend on derivatives of the metric