Invoking Continuum Hypothesis to prove the cardinality of a set To elaborate on the title, let me give an example.
Suppose I want to show that a certain $S$ has cardinality $\mathfrak{c}$.
For the sake of concreteness, suppose that $S = (0, 1) \subset \mathbb{R}$.
Is the following a valid proof?  


*

*I show that $|S| > \aleph_0$ by some argument. (Assume that I've done this correctly.) 

*Also, it is easy to see that $|S| \le \mathfrak{c}.$  

*Now, I know that the Continuum Hypothesis is independent ZF(C). What I have is an explicitly constructed set $S$. Thus, $|S|$ cannot be strictly between $\aleph_0$ and $\mathfrak{c}$. Ergo, $|S| = \mathfrak{c}.$

*(Otherwise, I would have an explicit construction that would disprove CH.)

*Thus, I have completed the proof without actually assuming CH but just the fact that it's independent of ZF(C).



On the same lines as this, what about the following proof which uses independence of choice from ZF?
I want to carry out the following proof without use of choice.
Consider $A = \mathbb{R}^\mathbb{R}$, $B = \{f \in A:f \text{ is continuous}\}$, and $C =  \{f \in A:f \text{ is discontinuous}\}$.
I want to show that $|C| = 2^\mathfrak{c}$.
Is the following correct?  


*

*I show that $|A|=2^\mathfrak{c},$ this is doable without choice.  

*I show that $|B| = \mathfrak{c} < 2^\mathfrak{c}$, again, without choice.

*It is clear that $|C| \le |A| = 2^\mathfrak{c}$.

*Now, I have $B \cup C = A$. If $|C| < 2^\mathfrak{c}$, then we would have that $A$ is an infinite set which is the union of two sets of strictly smaller cardinality.  

*Now, if we assume choice, then the above cannot happen. However, we have an explicit (constructive) example of these sets. So, if this were a counterexample, this construction would carry over in ZF+C as well.

*Thus, $|C| = 2^\mathfrak{c}$, without assuming choice.



One thing which I am implicitly assuming is that the cardinality of $|C|$ (resp. $|S|$) doesn't depend on whether or not I assume choice (resp. CH). Once again, my justification is that cardinality of these explicitly constructed sets should not depend on things like choice and CH, which are independent of ZF.
Of course, one more thing I'm assuming is that ZF is consistent.
 A: Your logic is wrong.
You are conflating truth and provability. While working in a given model of $\sf ZFC$ either $\sf CH$ is true or it's false, but one of them must be the case. Even if you don't know which one.
There are sets of reals that we can perfectly define, and their cardinality is either $\aleph_0$ or $\aleph_1$, regardless to the cardinality of the continuum, and so they may serve as a counterexample to the Continuum Hypothesis in some models and not in others.
In the case of $\sf AC$ your solution is entirely wrong. It is consistent without $\sf AC$ that every set which cannot be well-ordered is the union of two sets of smaller cardinality, and that in addition $\Bbb R$ cannot be well-ordered (so its power set cannot either).

Nevertheless, the two propositions that you're stating are true.


*

*In the case of $(0,1)$ we can define a bijection from $\Bbb R$ into $(0,1)$ (for example $x\mapsto 1/e^x$). Or we can define an injection from $\mathcal P(\Bbb N)$ into $(0,1)$.

*In the case of discontinuous functions, we can either define an injection from $\mathcal P(\Bbb R)$ into the set. Or we can appeal to an abstract theorem: If $|A|=|A|+|A|$, and $B\subseteq\mathcal P(A)$ such that $|A|=|B|$, then $|\mathcal P(A)|=|\mathcal P(A)\setminus B|$. The proof is not trivial, and you can find it elsewhere on this very site.
A: I'll only discuss the first question. As pointed out by Asaf, the argument is not correct, but something interesting can be said anyway.
There are a couple of issues. A key problem is with the idea of an "explicitly constructed" set. Indeed, for instance, there are explicitly constructed sets of reals that are uncountable and of size continuum in some models of set theory, uncountable and of size $\aleph_1<\frak c$ in some other models, and countable in yet others. An example is $\mathbb R^L$, the set of reals in Gödel's constructible universe $L$. This description is not typical of analysis, as it involves the metamathematical notion of $L$, but this does not matter much, as $\mathbb R^L$ is absolutely definable (meaning, there is a formula that defines it unambiguously in any model of set theory) in a $\Sigma^1_2$ way, that is, it is definable as the continuous image of the complement of the continuous image of a certain Borel set, and the two continuous functions and the Borel set involved in this description are also fairly explicit.
The set $\mathbb R^L$ is of course of size $\mathfrak c$ in $L$. In the model obtained by adding to $L$ $\aleph_2$ Cohen reals, it is uncountable of size $\aleph_1$, but in this model the reals have size $\aleph_2$. In the model obtained by collapsing $\aleph_1^L$ to size $\aleph_0$, it is countable.
This example suggests a second, perhaps subtle perhaps pedantic, problem, which is that you mention that in your argument you prove that $S$ is uncountable, but you did not specify in which theory this argument takes place (or, following the idea of models in the example above, in which model it is that $S$ is uncountable). One could remove the example I gave by insisting that the proof takes place in ZFC. However, this problem is not really serious, as there are standard ways of modifying the example to produce an explicit set that is always uncountable (meaning, provably in ZFC), but of size continuum in some models and of intermediate size in some others. A silly example would be to consider the set that is $\mathbb R^L$, should the reals of $L$ be uncountable, and $\mathbb R$ otherwise. 
Once could protest that the example above is not "explicit", but then the problem turns into having to formalize the notion of definability one actually means when one says things like "explicitly definable", and this is usually messy. However, it is here that one can say something interesting, at least if one is willing to replace ZFC by a stronger theory. First, one could say that "explicitly definable" means something like the $\Sigma^1_2$ definition of $\mathbb R^L$ (as opposed to the metamathematical one). This can be formalized in fairly general ways, so that we do not feel the notion of definability we settle for is "too restrictive". We could say, for instance, that "explicitly definable" means projective, that is, $\Sigma^1_n$ for some $n$ (so, a continuous image of the complement of the continuous image of ... of a Borel set). This is fairly generous, and it probably includes every set ever discussed in analysis. Or we could be even more generous and say that an "explicitly definable" set is one that belongs to $L(\mathbb R)$, the constructible closure of the reals. Any set in this class is definable from reals, $\mathbb R$, and ordinals, and there are many sets of reals captured by this notion that are significantly more complex than just about anything one ever encounters. To be slightly more concrete: $L(\mathbb R)$ admits a stratification that begins with $L_0(\mathbb R)=\mathbb R$ at the bottom, then continues with $L_1(\mathbb R)$, the collection of sets first-order definable with parameters in $L_0(\mathbb R)$ in an appropriate set-theoretic language, then proceeds to $L_2(\mathbb R)$, defined just as before but in $L_1(\mathbb R)$ in place of $L_0(\mathbb R)$, etc., all the way through the ordinals (taking unions at limit stages). All projective sets already appear in $L_1(\mathbb R)$, so this notion is vastly more generous than one probably needs.
Okay. Work now, not in ZFC, but in an extension obtained by adding to ZFC an appropriate large cardinal axiom. Explicitly, one could take 

ZFC+"there are $\omega$ Woodin cardinals, and a measurable cardinal larger than all of them". 

As far as large cardinals go, this is fairly tame by modern standards, and it is stronger than we need here. Provably in this theory, any set of reals in $L(\mathbb R)$ has the perfect set property. This means that any such set is either countable, or contains a perfect subset. In ZFC, any perfect set has size continuum. That is, your strategy actually succeeds (without needing to invoke CH): If you have an explicit set of reals (in the sense just indicated), and it is provably uncountable, then it must have size continuum, and the reason is "mathematical" (rather than the consequence of some metamathematical trick): the set contains a perfect subset.
Everything above is classical, and Jech's set theory book covers most of it. The result in the last paragraph appears in 

MR1074499 (92m:03087). Shelah, Saharon; Woodin, Hugh. Large cardinals imply that every reasonably definable set of reals is Lebesgue measurable. Israel J. Math. 70 (1990), no. 3, 381–394.

(That said, the Shelah-Woodin paper is not the most readable there is. This is the place where the notion of Woodin cardinal was isolated. A more modern exposition of stronger results appears in Neeman's chapter in the Handbook of set theory.)
At the cost of strengthening the large cardinal assumption, one can generalize even further the notion of definability to which the result applies. It is open what the precise limit of this process is (this is related to the complexity of the so-called universally Baire sets, and a careful discussion of it is more technical).
