Consider a plane $P$ that is normal to the charged planar sheet - for simplicity start with the $x$-$z$-plane. The charge density is mirror symmetric with respect to this plane and, in consequence, so is the electric field. You can do this with any plane normal to the charged plane, and as a result the field vector is normal to the charged planar sheet, i.e. the electric field only has an $x$-component. Further use either those mirror symmetries with respect to arbitrary normal planes, or simply the invariance of the charge density under translations in the $z$ or $y$-direction to find that the electric field cannot depend on $z$ or $y$.
We have thus found quite some info already just by exploiting symmetries of the system, and only now do we start into the calculation you present above, i.e. we use Gauß' law (generally valid, a statement that the electric flux through a surface is equal to the enclosed charge) for the surface $S$ you show above, i.e.
$$
\int_S \vec{D} \cdot \vec{n}\, \text{d}A = q\,.
$$
After using $\vec{D} = \varepsilon_0 \vec{E}$ and $q = A \sigma$, we turn our attention to the integral. Since, as we have found above, the electric field is normal to charged sheet, only on the faces labelled 1 and 2 is the scalar product $\vec{E} \cdot \vec{n}$ of electric field and surface normal non-zero. Even better: it is constant on that part of the surface because $\vec{E}$ does not depend on $y$ or $z$. Which means we can evaluate
$$
\int_S \vec{E} \cdot \vec{n}\, \text{d}A = 2 A \left|\vec{E}\right|\,.
$$
This is, in a bit more detail, what is going on in the derivation you are referencing. You see how important it is to exploit the symmetry, right?
Now, for the point charge issue. The electric potential - and thus the electric field as well - is calculated from the charge distribution in the whole of $\mathbb{R}^3$, not just a part in some Gaussian surface. As in (the solution to Poisson's equation)
$$
\varphi(\vec{x}) = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{x}')}{\left|\vec{x}-\vec{x}'\right|} \text{d}^3\vec{x}'\,,
$$
$$
\vec{E}(\vec{x}) = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{x}') (\vec{x}-\vec{x}')}{\left|\vec{x}-\vec{x}'\right|^3} \text{d}^3\vec{x}'\,.
$$
You can see this as an infinite superposition of the fields of point charges all over the charged plane.
Now, what we could have done instead of the "trick shot" involving Gauß' law is just to use the charge density of the infinite plane
$$
\rho(\vec{x}) = \sigma \delta(x)
$$
($x$ without arrow meaning the $x$-component of $\vec{x}$), plug this into the integral and also find the constant field from
$$
\vec{E}(\vec{x}) = \frac{\sigma}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\delta(x) (\vec{x}-\vec{x}')}{\left|\vec{x}-\vec{x}'\right|^3} \text{d}^3\vec{x}'\,.
$$
(Which of course is a bit more tricky, an exercise in evaluating integrals... ;) )