Let $a=\sqrt{4+\sqrt{5-a}}, b=\sqrt{4+\sqrt{5+b}}, c=\sqrt{4-\sqrt{5-c}}$ and $d=\sqrt{4-\sqrt{5+d}}$.

Calculate $a b c d$.
Since we are dealing with polynomials, let's honor them by getting rid of those nasty roots. $a, c, -b, -d$ are roots of $X^{4}-8 X^{2}+X+11$. So $a b c d = b d \times(-a) \times(-c)=11$ and we are done... no! To be sure we have the four roots of the polynomial (the product of the opposites of which gives the constant term) without any multiplicity issues, it would be good to show that these four roots are distinct. $a, c$ are positive, $-b, -d$ are negative. If $a=c$, $\sqrt{5-a}=-\sqrt{5-a}=0$. So $a=5$, and from the statement, $a=\sqrt{4}=2$, contradiction. Therefore, $a \neq c, b \neq d$ similarly, and we have the desired result.