Equivalence class on real numbers Call two real numbers equivalent if their binary decimal expansion differ in a finite amount of places, if S is a set which contains an element of every equivalence class, must S contain an interval?
How to show that every interval contains an (uncountable number of?) element of every equivalence class? 

Added part: We produce a bounded set $S$ that contains a member of every equivalence class but does not contain an interval. Every equivalence class meets $[0,1]$, since for any $x$, we can, by making a finite number of changes to the bits of $x$, produce an $x'\in [0,1]$.
Use the Axiom of Choice to select $S\subset [0,1]$ such that $S$ contains precisely one member of each equivalence class.  Any two dyadic rationals (expressed in ultimately $0$'s form) belong to the same equivalence class, so $S$ contains exactly one dyadic rational.  Since the dyadic rationals are dense in the reals, this means that $S$ cannot contain an interval of positive length.
(End of added part)
Every non-empty interval $I$ contains a member of every equivalence class.
For let $I$ be a (finite) interval, and let $a$ be its midpoint. Suppose that $I$ has length $\ge 2\times 2^{-n}$.  Let $x$ be any real number.  By changing the initial bits of $x$ so that they match the initial bits of $a$, up to $n$ places after the "decimal" point, we can produce an $x'$ equivalent to $x$ which is at distance less than $2^{-n}$ from $a$.
Edit: The question has changed to ask whether every interval contains an uncountable number of members from every equivalence class.  Minor modification of the first paragraph shows that every interval contains a countably infinite number of members from every equivalence class. As Asaf Karagila points out, one cannot get more, since every equivalence class is itself countable. (The set of places where there is "change" can be identified with a finite subset of the integers, and $\mathbb{N}$ has only countably many finite subsets.)
