You could prove that by appealing to "any closed and bounded subset of the real numbers is compact" or directly from the definition:
Let [itex]\{U_\alpha\}[/itex], for [itex]\alpha[itex]in some index set, be an open cover for this set. Since 1 is in the set, 1 is in at least one of those- relabel the sets, if necessary, so that set is called [itex]U_1[/itex]. Since [itex]U_1[/itex] is open, there exist a point p, in that set, and a number [itex]\delta> 0[/itex] such that [itex]N_\delta(p)= \{x | |x- p|<\delta\}[/itex] is in [itex]U_1[/itex]. It then follows that if [itex]n> 1/\delta[/itex] [itex]|1- (n-1)/n|= 1/n< \delta[/itex] so that all except a finite number of the fractions in the sequence are in [itex]U_1[/itex]. Every number in the sequence, [itex](n-1)/n[/itex] for n less than that might be in as separate set in the original sequence but there are only a finite number of them.[/itex][/itex]