The requirement $3-x \geq 0$ is wrong. Whoever wrote this solution was probably thinking that $|x-1|$ is always non-negative, and $|x-1| \geq 3-x$, so $3-x$ is non-negative, but that reasoning is scrambled.
(If the inequality in the problem went in the other direction, i.e. if the original question were $|x-1| \leq 3-x$, then you could reason that $|x-1| \geq 0 \implies 3-x \geq 0$, but that would be a different problem.)
In fact you can just take any $x>3$ and see that indeed it is a solution to the inequality. For example, with $x=4$ we have $|4-1| \geq 3-4$, i.e. $3 \geq -1$, which is obviously true.
Edited to add: A better approach to solving the absolute value inequality is to break the problem into two cases.
- Case 1: If $x-1$ is non-negative, then $|x-1| = x-1$, and so we have the compound inequality $x \geq 1$ and $x-1 \geq 3-x$. Solve and find $x \geq 2$.
- Case 2: If $x-1$ is negative, then $|x-1| = 1-x$, and so we have the compound inequality $x < 1$ and $1-x \geq 3-x$. Solve and find this has no solutions.
So the solution is $x \geq 2$, full stop.