How would you solve this problem for real $x$? $$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
It can be easily shown that both equations $$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$ and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$ have the same real solution, i.e. the golden ratio, $$x = \frac {\sqrt 5 + 1}2=\phi.$$
The main question is derived by combining (1) and (2) into one equation. The final solution is still the same, but the approach might not be as straightforward.
With some difficulty.
You can turn the equation into a polynomial through lots of squaring and rearranging, starting with:
$\begin{eqnarray} \sqrt{2+\sqrt{2-x}} & = & \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} \\ & = & \sqrt{\frac{x - 1}{x}}\left(\sqrt{x + 1} + 1\right) \\ 2 + \sqrt{2 - x} & = & \frac{x - 1}{x} \left( x + 2 + 2 \sqrt{x + 1} \right) \\ \frac{(x-1)(x+2)}{x} - 2 & = & \sqrt{2 - x} - 2 \frac{x - 1}{x}\sqrt{x + 1} \\ x^2 - x - 2 & = & x\sqrt{2 - x} - 2(x-1)\sqrt{x+1} \end{eqnarray}$
After a few more rounds, you should get to something vaguely resembling $x^8-10x^7+39x^6-22x^5-63x^4+32x^3+32x^2 = 0$. You can divide out $x^2$ (which is clearly not a solution to the original equation), and then you can use the knowledge that $\varphi$ is one of the roots to divide out the quadratic $x^2 - x - 1$ (since the polynomial has integer coefficients, you can deduce that $-\varphi^{-1}$ is also a root, which is an erroneous solution introduced in one of the squaring steps).
At this point, you're left with $x^4-9x^3+31x^2-32 = 0$. This quartic has two complex roots and two real roots, but actually figuring them out analytically is what mathematicians call "damn hard", and is commonly achieved via the method known as "chuck it into Mathematica".
(The following is about the other roots in ConMan's answer, but it was too long for a comment.)
Courtesy WA, the original equation can be brought to polynomial form with groebnerBasis[ {u - v - w, x - 2 + (u^2 - 2)^2, v^2 x - x^2 + 1, w^2 x - x + 1}, {x}, {u, v, w} ] which returns and factors:
$$
x^6 - 10 x^5 + 39 x^4 - 22 x^3 - 63 x^2 + 32 x + 32 = (x^2 - x - 1) (x^4 - 9 x^3 + 31 x^2 - 32)
$$
The positive root $\,\varphi\,$ of the quadratic and the negative real root $\,\simeq -0.8959\,$ of the quartic satisfy the original equation $\,\sqrt{2 + \sqrt{2-x}} = x = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.
The negative root $\,-\dfrac{1}{\varphi}\,$ of the quadratic satisfies
$\,\sqrt{2 - \sqrt{2-x}} = -x = - \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.
The positive real root $\,\simeq 1.2197\,$ of the quartic satisfies
$\,\sqrt{2 - \sqrt{2-x}} = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.
The two complex roots $\,\simeq 4.3381 \pm 3.2349 i\,$ of the quartic satisfy
$\,\sqrt{2 - \sqrt{2-x}} = \sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\,$.
Let $$f(x)=\sqrt{2+\sqrt{2-x}}$$ and $$g(x)=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
See the fig below where $y=f(x)$ blue and $y=g(x)$ green and $y=x$ (red).
The solution of $x=f(x)$ is given by the intersection of red and blue lines at $x=1.6180...=\frac{1+\sqrt{5}}{2}$ (the same golden number)
The root of $x=g(x)$ is also given intersection of red and green at the golden number.
But the roots of the equation $f(x)=g(x)$ is given by two points of intersection of blue and green lines, one at the golden number and the other at $x=-0.8959...$
Also it turns out that the common root of $x=f(x), x=g(x), f(x)=g(x)$ is at $\frac{1+\sqrt{5}}{2}$ and not at $\frac{\sqrt{5}-1}{2}.$
