To understand this problem, you must be familiar with the
notions of countable and uncountable sets.
A countable infinite set is one whose members can be placed into
one-to-one correspondence with the natural numbers.
The members of an uncountable set, however, cannot be placed into
one-to-one correspondence with the natural numbers.
Cantor proved that the real numbers are uncountable, before
going insane.

A collection of sets is known as a "chain" if it has the property
that for any two sets in the collection, one of them is a proper
subset of the other.

The problem: Is it possible to construct an uncountable chain of subsets
of the natural numbers?

It is possible, and we'll show how to construct an uncountable chain of subsets of the natural numbers.

First we define a function f(n) from the natural numbers
to the set of positive rationals.
The idea is make a list of the positive rationals by sorting them first
by the sum of numerator plus denominator, and then by numerator.
We can skip any values for which the numerator and denominator have
a common factor.

The first 14 values of this function are then:

f(1) = 1/1

f(2) = 1/2

f(3) = 2/1

f(4) = 1/3

f(5) = 3/1

f(6) = 1/4

f(7) = 2/3

f(8) = 3/2

f(9) = 4/1

f(10) = 1/5

f(11) = 5/1

f(12) = 1/6

f(13) = 2/5

f(14) = 3/4

etc.

Now we can define our chain of set as follows. Each set is associated
with a real number r > 1:

S_{r} = { n | f(n) <= r }

These are clearly uncountable, since they can be placed into one to one
correspondence with the real numbers greater than or equal to one,
and Cantor showed that those numbers are uncountable.

It's also clear that S_{t} is a subset of S_{u} whenever
1 < t < u.

All that's left is to show that S_{t} is a **proper** subset of
S_{u} whenever 1 < t < u.
But that's easy!
If t < u, just choose as denominator d any integer greater than 1/(u-t),
and there will be a multiple m of 1/d such that t < m/d < u.
So m/d is in S_{u} but not in S_{t}.
And we're done!