Have you studied the famous Cantor Set ?
Suppose I have a closed, bounded, non-degenerate interval .
I have a set X initially containing the points a and b.
My first action is to find the midpoint of I, call it c, and add c to X.
I now have two new intervals and .
My next action is to find the midpoints of and , lets call them and add them to X.
I now have four intervals.... and I continue this procedure until I can subdivide no more.
Does X contain all points in the original interval I? Or stated differently, is there a point in I that is not in X?
Ignoring the point about "subdivide no more", isn't it obvious that the set of all subintervals does include every point in the original interval? Ever point, p, in [a, b] satisfies . For any point c, either or or both. So if x is in [a, b], either x is in [a, c] or x is in [c,b]. In either case, x is in [a, c]U[c, b]. The general proof can be done by induction on the number of halvings.
Then we form a partition of $[a,b]$ with that new set of points and find the midpoints of each new subinterval.
So after each stage we have a set of a finite collection.