electrostatics - Why can charges outside be ignored in Gauss's Law?

In MIT's 8.02 course, it is shown in lecture 3 that we can derive Gauss's Law from Coulomb's to get

$ \phi = \oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_{0}} $

However, in the lecture, it was assumed that there were no charges outside the surface. Later, Gauss's Law was used and the charges outside the sphere were ignored.

I've been thinking about it and came across Can someone give an intuitive way of understanding why Gauss's law holds?.

The first answer by user levitopher helped a bit. I think that the lecture's argument still holds when there are charges outside because $ \vec{E} $ is added vectorially.

Suppose we have $ Q_{enc} $ in a sphere and $ Q_{out} $ outside the sphere. Then,

By definition, $ \phi = \oint (\vec{E_{Q_{enc}}} + \vec{E_{Q_{out}}}) \cdot \vec{dA} $
$ = \oint \vec{E_{Q_{enc}}} \cdot \vec{dA} + \oint \vec{E_{Q_{out}}} \cdot \vec{dA}$
$ = \oint \vec{E_{Q_{enc}}} \cdot \vec{dA} + 0 $
$ = \frac{Q_{enc}}{\epsilon_{0}} $

This assumes that $ \oint \vec{E_{Q_{out}}} \cdot \vec{dA} = 0 $ because any field lines that go in must come out (hand wavy but I'm just going to accept it for now).

I'd just like to know if this argument is sound.