Calculating the pH of a highly dilute solution of HCl
At first you have to think of where the protons come from. That is as you mentioned it, hydrochloric acid (for every chloride ion in solution there has to be a proton) and water (for every hydroxide ion from the autoprotolysis there also has to be a proton). $$\ce{[H+] = [OH-] + [Cl-]}$$

The concentration for the hydroxide ions can be derived from the autoprotolysis of water, as I've said before.
$$k_\mathrm w = \ce{[H+]~[OH-]}
\Rightarrow \ce{[OH-]}=\frac{k_\mathrm w}{\ce{[H+]}}$$

Using some funny math, one can find out, that the concentration of the chloride ions in solution can be described the following eqation. Recognizing shortly after that the hard way was not necessary yields to the simplification that hydrochloric acid is completely dissociated.
$$\ce{[Cl-]} = \frac{\ce{[HCl]0}~k_\mathrm a}{\ce{[H+]} + k_\mathrm a} \xrightarrow{k_\mathrm a \gg \ce{[H+]~~}} \frac{\ce{[HCl]0}~k_\mathrm a}{k_\mathrm a} = \ce{[HCl]0}~(=c_0)$$

This changes the first equation to be like
$$\ce{[H+]} = \frac{k_\mathrm w}{\ce{[H+]}} + c_0$$

Rearranging everything gives you a second order equation in $\ce{[H+]}~(=x)$
$$x^2-c_0~x-k_\mathrm w = 0$$
Wherefrom you get two solutions (from which only the positive one makes sense):
$$x = \frac{c_0}{2} + \sqrt{\frac{c_0^2}{4}+k_\mathrm w}$$

Let's compare this solution to your mentioned version. The following graph shows you the difference between both assumptions as $$\operatorname{abs}(f_1(c)-f_2(c))$$

$\hskip3.5cm$

A maximum difference of 0.1 pH between the easier and the "more complex" version is acceptable.