$\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational

In your proof, after $6-2x\sqrt{6}+x^2=5+2\sqrt{6}$ we have that $$x^2+1=2(x+1)\sqrt{6}$$ If $x=-1$ then, from the above equation, it follows that $2=0$. Therefore $x$ is a rational number different from $-1$. After dividing by $2(x+1)\not=0$ we get $$\sqrt{6}=\frac{x^2+1}{2(x+1)}\in \mathbb{Q}.$$ Contradiction! Hence $x=\sqrt{6}-\sqrt{2}-\sqrt{3}$ is not a rational number.

$\sqrt{6}-\sqrt{2}-\sqrt{3}$ is a root of $x^4 - 22 x^2 - 48 x - 23$.

By the rational root theorem, $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is either irrational or an integer.

But $$ 1.4 < \sqrt 2 < 1.5 \\ 1.7 < \sqrt 3 < 1.8 \\ 2.4 < \sqrt 6 < 2.5 \\ $$ imply $$ -0.9 < \sqrt{6}-\sqrt{2}-\sqrt{3} <-0.6 $$ and so $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is not an integer. Therefore, it is irrational.