With conditionals, I believe that the most generally applied method of proof consists in assuming the left part of the conditional and then deriving the right part of the conditional.
Now, for a conditional with a disjunction in the consequent, as Mauro's comments suggest, we first want to ascertain the nature of the disjunction. Is it an inclusive disjunction or an exclusive disjunction? In other words, can both of the disjuncts hold true or can only one of the disjuncts hold true?
If both of the disjuncts X, Y can hold true, we have some options. One option lies in proving that one of the disjuncts does hold. Consequently, the disjunction will follow. Another option starts by assuming the negation of the disjunction. Since ($\lnot$(X$\lor$Y)$\rightarrow$($\lnot$X$\land$$\lnot$Y)), ($\lnot$X$\land$$\lnot$Y) follows. Then we can consider $\lnot$X and $\lnot$Y separately in turn. If either $\lnot$X or $\lnot$Y more immediately leads to a contradiction, then so does $\lnot$(X$\lor$Y) and (X$\lor$Y) follows as true.
So, for your example suppose that $\lnot$ ((x < 2) $\lor$ (x >= 3)) under the scope of the left part of your conditional. Thus, ($\lnot$(x<2) $\land$ $\lnot$(x >= 3)). So, we can have any x such that x > 2. But, suppose that x = 2.1. Then from the left part of your conditional we obtain (2.1 / .1) <= 3. But, 21 <= 3 is false and thus we have a contradiction. Consequently, $\lnot$ ((x < 2) $\lor$ (x >= 3)) is false and ((x < 2) $\lor$ (x >= 3)) is true.
Therefore, your conditional follows.
You might not find that convincing. Well, if so or if not so, then I don't find it convincing either. But, I don't know of any logical errors. And proofs don't have to convince to work.