Given the set A={x|-5<x^3<5} and the set B={-3,-1,0,2,3}, how to find the intersection of sets A and B?
I tried the following method myself, but how come it doesn't achieve the goal?
A = {Reduce[{-5 < x^3 < 5}, x, Reals]} B = {-3, -1, 0, 2, 3} Intersection[A, B] 
If I understand Intersection it works on discrete lists (typically List, but any head it seems). My best guess is that you want an Interval.
intervalA = Interval[{A[[1, 1]], A[[1, -1]]}]; Select[B, IntervalMemberQ[intervalA]] (* {-1, 0} *) @lericr already explained this very nicely and gave a proper answer. I just want to provide an answer with minimal modification of the original question for pedagogical reasons.
Create a list with the output of Reduce (PROPERLY) and then use Intersection
It works in the following manner
A = x /. List@ToRules@Reduce[{-5 < x^3 < 5}, x, Integers]; B = {-3, -1, 0, 2, 3}; Intersection[A, B] 
Intersection has to be used. Frankly your solution is the proper one in my opinion. $\endgroup$ f[u_] := Reduce[RealAbs[x^3] < 5, x] /. x -> u; b = {-3, -1, 0, 2, 3} Select[f@# == True &][b] yields {-1,0}
NumberLinePlot[{f[x], b}, {x, -4, 4}]
set = Range[-10, 10]; A = Pick[set, -5 < #^3 < 5 & /@ set] B = {-3, -1, 0, 2, 3}; Intersection[A, B] {-1, 0, 1}
{-1, 0}
Direct result from Reduce:
Reduce[-5 < x^3 < 5 && AnyTrue[{-3, -1, 0, 2, 3}, x == # &], x] x == -1 || x == 0
B = {-3, -1, 0, 2, 3} Select[B, -5 < #^3 < 5 &] 
Intersectioncarefully, don't guess its usage, it's just not for the task you're imagining. (-1) $\endgroup$