The first part of the problem was to solve for $a=3$ implies $x\in [0,2]$. This makes me think that I will somehow use this. I try to find the solution for the second part of the problem by squaring the inequality. I get $2\sqrt{(-{x}^{2}+2x)}\ge a-2$.

This is equal to two systems:

$\{\begin{array}{l}a<2\\ x\in [0,2]\end{array}\phantom{\rule{1em}{0ex}}{\textstyle \text{and}}\phantom{\rule{1em}{0ex}}\{\begin{array}{l}a\ge 2\\ 4{x}^{2}-8x+{a}^{2}-4a+2\ge 0\end{array}$

What do I do know and is this actually correct?

This is equal to two systems:

$\{\begin{array}{l}a<2\\ x\in [0,2]\end{array}\phantom{\rule{1em}{0ex}}{\textstyle \text{and}}\phantom{\rule{1em}{0ex}}\{\begin{array}{l}a\ge 2\\ 4{x}^{2}-8x+{a}^{2}-4a+2\ge 0\end{array}$

What do I do know and is this actually correct?