Question about partitions in intervals of the real numbers. I have to prove the following:

Let $ \mathcal{D} $ be a partition of $ \mathbb{R} $ in intervals of any
  kind, except intervals containing a single element. Prove
  $ \mathcal{D} $ is countable.

Now if $\mathcal{D}$ is finite the result is trivial; my problem is when it is infinite. In this case there is an inyective function $ f: \mathbb{N} \rightarrow \mathcal{D}, $ so $ \left| \mathbb{N} \right| \leq \left| \mathcal{D} \right|. $ But I don't know how to construct an injective function in the other direction to, then, apply the Cantor-Bernstein Theorem. In fact, I'm not even sure this is the right way to prove the result. 
I would like to read some suggestions on how to tackle this problem.
 A: Notice that $\mathbb{Q} \subseteq \mathbb{R}$ and $\left| \mathbb{Q} \right| = \left| \mathbb{N} \right|.$ Now, every interval, other than a single element interval, contains a rational number, so you can map every interval in your partition to a rational number contained in that particular interval. Since the intervals in your partition are disjoint you don't have to worry about mapping two intervals to the same rational, so the map is injective. Clearly the map is also surjective, so you have a bijection between your partition to $\mathbb{Q},$ and hence to $\mathbb{N}$ as desired.
You should obviously write this argument a little nicely ;) 

UPDATE: I think that in this proof we are using the Axiom of Choice when we say 

You can map every interval in your partition to a rational number
  contained in that particular interval.

However I don't think this is a serious problem.
UPDATE: Oops! This map is not surjective, but is injective, so you have one inequality. The other inequality is the one you obtained. 
A: HINT: There is a countable dense subset to the real numbers. Recall the definition of being dense.
