How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides

\begin{eqnarray}\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &=& \sqrt{4+2\sqrt3 \over 2}-\sqrt{4-2\sqrt3 \over 2}\\ &=&\sqrt{(\sqrt{3} +1)^2 \over 2}-\sqrt{(\sqrt{3} -1)^2\over 2}\\ &=&{\sqrt{3} +1 \over \sqrt{2}}-{\sqrt{3} -1\over \sqrt{2}}\\ &=&\sqrt2 \end{eqnarray}

Let $a=\sqrt{2+\sqrt3}\,$, $b = \sqrt{2-\sqrt3}\,$, then:

$$\require{cancel} a^2+b^2 = 2+\bcancel{\sqrt{3}}+2 - \bcancel{\sqrt{3}} = 4 \\ ab = \sqrt{(2+\sqrt3)(2-\sqrt3)} = \sqrt{2^2 - (\sqrt{3})^2} = \sqrt{4-3} = \sqrt{1} = 1 $$

It follows that: $$(a-b)^2 = a^2+b^2-2ab = 4 - 2 \cdot 1 = 2$$

Since $\sqrt{2+\sqrt3} \gt \sqrt{2-\sqrt3}\,$, $a-b \gt 0$ must be the positive root, so $a-b=\sqrt{2}\,$.

First of all we're not trying to find a solution of the equation here, what you are suggesting is to prove that $\mathrm{lhs} =\sqrt2 $ To do so we square the lhs (first read it fully) and we get $2$. So lhs would be $\sqrt2$ or $-\sqrt2$.

Now we observe the fact that lhs was positive initially ( as $ 2+\sqrt3 > 2-\sqrt3 $) hence lhs would take the positive value ie. $ +\sqrt2$, which is equal to rhs.

So I think it can be solved by observation and easy maths.