To anyone else wanting to comment on this thread, it helps to view Krizalid's .pdf

here, and see the context on page 4.

LHS = left hand side, RHS = right hand side, used below in reference to the inequality you're trying to solve.

So Krizalid isn't placing a restriction $\displaystyle x-2\le0$ but rather is putting this forward as a candidate solution set. Since the radicand is defined for this entire interval, and because the LHS is therefore nonnegative and the RHS nonpositive for this interval, the LHS is greater than or equal to the RHS and the inequality holds. Thus $\displaystyle (-\infty, 2]$ is part of the solution set.

You said x=3 doesn't satisfy the inequality but actually it does.. see the steps that occur after both sides are squared.

Does this answer your question?