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?

The number $\sqrt{1-x}-\sqrt{x}$ can be negative and, in this case, the inequality doesn't hold. Squaring both sides can be done only if both sides are nonnegative. Thus your inequality becomes
$$
\begin{cases}
0\le x\le 1 \\[8px]
\sqrt{1-x}-\sqrt{x}\ge0 \\[4px]
1-x-2\sqrt{x(1-x)}+x>\dfrac{1}{3}
\end{cases}
$$
The second inequality becomes $1-x\ge x$, so $x\le1/2$. The third inequality is
$$
\sqrt{x(1-x)}<\frac{1}{3}
$$
so $9x^2-9x+1>0$ that has the solutions
$$
x<\frac{3-\sqrt{5}}{6}\qquad\text{or}\qquad x>\frac{3+\sqrt{5}}{6}
$$
that should be combined with $0\le x\le 1/2$. Thus the solution set of the original equation is
$$
\left[0,\frac{3-\sqrt{5}}{6}\right)
$$
because $(3+\sqrt{5})/6>1/2$.

A different strategy is to write the inequality as
$$
\sqrt{3}\sqrt{1-x}>1+\sqrt{3}\sqrt{x}
$$
With the constraint $0\le x\le1$, this becomes
$$
3-3x>1+2\sqrt{3}\sqrt{x}+3x
$$
or
$$
3x+\sqrt{3}\sqrt{x}-1<0
$$
Setting $t=\sqrt{x}$ (with $0\le t\le1$), this is quadratic: $3t^2+\sqrt{3}t-1<0$. So we have the solution set
\begin{cases}
0\le t \le 1 \\[8px]
\dfrac{-\sqrt{3}-\sqrt{15}}{6}<t<\dfrac{-\sqrt{3}+\sqrt{15}}{6}
\end{cases}
that becomes
$$
0\le t<\dfrac{-\sqrt{3}+\sqrt{15}}{6}
$$
and therefore
$$
0\le x<\left(\dfrac{-\sqrt{3}+\sqrt{15}}{6}\right)^2=\frac{3-\sqrt{5}}{6}
$$
