Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$? Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which yields:
$$\frac{1}{3} > \sqrt{x}\sqrt{1 -x}.$$ 
Squaring it again we get inequality:
$$9x^2 - 9x + 1 > 0,$$
solution to which is a domain $[0, \frac{3 - \sqrt{5}}{6}) \cup (\frac{3 + \sqrt{5}}{6}, 1]$. But, solution to original inequality is just $[0, \frac{3 - \sqrt{5}}{6})$. What am I omitting/where I'm doing mistakes?

Observe first that if must be $\;0\le x\le 1\;$ , so the given interval is within this constraint. Second, square:
$$\sqrt{1-x}-\sqrt x\ge\frac1{\sqrt3}\implies1-x-2\sqrt{x(1-x)}+x\ge\frac13\implies$$
$$2\sqrt{x(1-x)}\le\frac23\implies x(1-x)\le\frac19\implies 9x^2-9x+1\ge0\implies$$
The last quadratic's roots are given by
$$x_{1,2}=\frac{9\pm\sqrt{45}}{18}=\frac{3\pm\sqrt5}6\implies$$
$$ 9x^2-9x+1\ge0\iff \left(x-\frac{3-\sqrt5}6\right)\left(x-\frac{3+\sqrt5}6\right)\ge0\iff$$
$$x\le\frac{3-\sqrt5}6\;\;\text{or}\;\;x\ge\frac{3+\sqrt5}6\;\;\;\color{red}{(***)}$$
and  we get what we want from the leftmost inequality.
Added on request: If we take
$$f(x)=\sqrt{1-x}-\sqrt x\implies f'(x)=-\frac1{2\sqrt{1-x}}-\frac1{2\sqrt x}=-\frac12\left(\frac1{\sqrt{1-x}}+\frac1{\sqrt x}\right)<0\implies$$
$\;f\;$ is monotone descending, and since 
$$f\left(\frac{3-\sqrt5}6\right)=\sqrt{1-\frac{3-\sqrt5}6}-\sqrt\frac{3-\sqrt5}6=\frac1{\sqrt6}\left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)\le\frac1{\sqrt3}\;(*)$$
since
$$(*)\iff \left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)^2\le\left(\sqrt\frac63\right)^2\iff6-4\le2\;\;\color{green}\checkmark$$
so in $\;\color{red}{(***)}\;$ only the left inequality applies.
