Apparent contradiction regarding countable subsets of real numbers. So I'm working through this HW problem and I think I did everything correctly but I get to a contradiction.
We start with $M$ which is a set of positive real numbers such that the sum of any subset of $M$ is smaller or equal to $1$. 
I then prove that $M_r = \{x\in M | x > \frac{1}{r}\}$ has at most $r$ elements and is therefore finite.
I then prove that $M=\bigcup_{n=1}^{\infty}M_r$. 
But here's the contradiction. The union is countable and $M_r$ is finite and using this theorem:

The union of countably many finite or countable sets is again finite
  or countable

We get that $M$ is finite. But $M$ does not need to be finite. There are many examples of $M$ being infinite. 
So where am I going wrong in my reasoning because there seems like a contradiction?! Am I maybe misunderstanding what the theorem says?
 A: I think you're confusing the theorem by breaking it up into two parts:
The union of countably many finite sets is finite
The union of countably many countable sets is countable
However, the theorem is one statement saying that the union of countably many (finite OR countable) sets is (finite OR countable).
So, even if I have a countable union of finite sets, the result of course need not be finite. For example, $\mathbb{N}$ itself is a countable union of finite sets, namely 
$$\bigcup_{n=1}^{\infty} \{n\}.$$
But then again, if I have a countable union of finite sets, the result could be finite.
For example, if only finitely many sets are non-empty, then the union is finite (in essence, we have a finite union of finite sets, which is then finite). Or, I could have the union of a countable number of sets which all contain the elements $\{1, 2, 3\}$. Well, their union is just $\{1, 2, 3\}$, which is finite.
That said, the result that a countable union of countable sets is still countable is certainly the most remarkable result of the theorem. In that regard, I think it would be more meaningful to state the theorem just slightly differently as
A countable union of at most countable sets is at most countable.

Or even
An at most countable union of at most countable sets is at most countable.

Where "at most countable" is the same as saying "either finite or countable" (i.e., no larger than countable).
A: Your statement $M=\cup_{n=1}^{\infty}  M_r$ is either false or meaningless. 
For each $q\in Q^+$ the set $M_q$ is finite.  Every $M_q$ is a subset of $M. $ And  $x\in M\implies \exists q\in (0,x)\cap Q^+)\implies x\in M_q.$ So $M\subset \cup_{q\in Q^+}M_q\subset M.$ So $M=\cup_{q\in Q^+}M_q$ is a countable union of finite sets, so $M$ is countable.
A: It might be worth considering: let $S = \Bbb{Q} \cap (0,1]$ and take $M_1 = \{1\}$, $M_2 = \{ 1/2, 2/2=1 \}$, $M_3 = \{ 1/3, 2/3, 3/3=1 \}$, ... .  The collection $\{M_j\}_{j \in \Bbb{Z}_{\geq 1}}$ is a countable collection of finite sets and each element of S is in at least one $M_j$.  (Actually, each element of $S$ appears in infinitely many of the $S_j$.)  So the countable union of finite sets, $M = \bigcup_{j \in \Bbb{Z}_{\geq 1}} M_j $, is countably infinite (and has to be -- any way of assembling the rationals in the unit interval must end up with infinitely many of them).
