Don't use a calculator.

In general, let $S = \sqrt{a + \sqrt{b}} - \sqrt{a - \sqrt{b}}$ for positive reals $a,b.$

Now $S^{2} = (a + \sqrt{b}) + (a - \sqrt{b}) - 2\sqrt{(a + \sqrt{b})(a - \sqrt{b})} = 2(a - \sqrt{a^{2} - b}).$

Assuming that $a \ge \sqrt{b}$ it is clear that $S \ge 0,$ and so $S = \sqrt{2(a - \sqrt{a^{2} - b})}.$

In this case $a = 23, b = 408,$ and thus

$S = \sqrt{2(23 - \sqrt{529 - 408})} = \sqrt{2(23 - 11)} = \sqrt{24} = \boxed{2\sqrt{6}}.$

$\begin{array}{l} \sqrt {23 + \sqrt {408} } - \sqrt {23 - \sqrt {408} } = \\ \mathop = \limits_{(1)} \sqrt {17} + \sqrt 6 - \left( {\sqrt {17} - \sqrt 6 } \right)\\ = \sqrt {17} + \sqrt 6 - \sqrt {17} + \sqrt 6 \\ = 2\sqrt 6 \\ \\ (1)\sqrt {A \pm \sqrt B } = \sqrt {\frac{{A + C}}{2}} \pm \sqrt {\frac{{A - C}}{2}} \quad ,\quad {C^2} = {A^2} - {B^2} \end{array}$

Let $\space \begin{cases} (\sqrt{a}+\sqrt{b})^2 = 23 + \sqrt{408} \\ (\sqrt{a}-\sqrt{b})^2 = 23 - \sqrt{408} \end{cases} \quad \text{where } a > b$

$\Rightarrow \sqrt{23+\sqrt{408}} - \sqrt{23+\sqrt{408}} = (\sqrt{a}+\sqrt{b}) - (\sqrt{a} - \sqrt{b}) = 2\sqrt{b}$

$\begin{aligned} (\sqrt{a}+\sqrt{b})^2 & = a+b+2\sqrt{ab} = 23 + \sqrt{408} \end{aligned}$

$\Rightarrow \begin{cases} a + b = 23 & \Rightarrow a = 23 - b \\ 2\sqrt{ab} = \sqrt{408} & \Rightarrow ab = 102 \quad \Rightarrow (23-b)b = 102 \\ & \Rightarrow b^2 - 23b + 102 = 0 \quad \Rightarrow (b-6)(b-17) = 0 \end{cases}$

Since $\space a > b \quad \Rightarrow a = 17 \quad \Rightarrow b = 6$

$\Rightarrow \sqrt{23+\sqrt{408}} - \sqrt{23+\sqrt{408}} = 2\sqrt{b} = \boxed{2\sqrt{6}}$