How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:

$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$

Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
However I know you can't square a side during proof because it generates an extraneous solution. So: how do you go about this proof without squaring both sides? Or can my method be made valid if I do this:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
$$...=...$$
$$2=2$$
$$\lvert\sqrt2\rvert=\lvert\sqrt2\rvert$$
$$\sqrt2=\sqrt2\text{ hence proved.}$$
Cheers in advance :)

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\begin{align}
   \sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= x \tag{A}\\
   \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)\bigg(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\bigg) 
   &= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\
   (2+\sqrt 3)-(2-\sqrt 3) &= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\
   \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x &= 2\sqrt 3 \\
   \sqrt{2+\sqrt3} + \sqrt{2-\sqrt3} &= \dfrac{2\sqrt 3}{x} \tag{B}\\
   2\sqrt{2+\sqrt 3}  &= \dfrac{2\sqrt 3}{x} + x \tag{B+A}\\
   \sqrt{2+\sqrt 3}  &= \dfrac{\sqrt 3}{x} + \dfrac x2 \tag{C}\\
   2\sqrt{2-\sqrt 3}  &= \dfrac{2\sqrt 3}{x} - x \tag{B-A}\\
   \sqrt{2-\sqrt 3}  &= \dfrac{\sqrt 3}{x} - \dfrac x2 \tag{D}\\
   1 &= \dfrac{3}{x^2} - \dfrac{x^2}{4} \tag{CD} \\
   4x^2 &= 12 - x^4 \\
   x^4 + 4x^2 - 12 &= 0 \\
   (x^2 - 2)(x^2 + 6) &= 0 \\
   x &= \sqrt 2 \\
   \sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= \sqrt 2
\end{align}
