Minimal Generalized Continuum Hypothesis & Axiom of Choice It is well known that working in the frame of $\text{ZF}$, the Generalized Continuum Hypothesis ($\text{GCH}$) implies the Axiom of Choice ($\text{AC}$), i.e. $\text{ZF}+\text{GCH}\vdash \text{AC}$. 
But if we consider $\text{GCH}$ as a theory with ordinal many statements like  $\text{GCH}=\{\text{CH}_{\alpha}~|~\alpha\in \text{Ord}\}$ such that $\text{CH}_{\alpha}$ is the statement $2^{\aleph_{\alpha}}=\aleph_{\alpha +1}$, then there is a natural question as follows:
Is assuming all of these strong statements really necessary to prove a weak proposition like Axiom of Choice? 
Precisely: 
Question (1): Is there a class $\text{C}\subsetneq \text{Ord}$ such that: 
(a) The assumption $\{\text{CH}_{\alpha}~|~\alpha\in \text{C}\}$ is strictly weaker than the assumption $\{\text{CH}_{\alpha}~|~\alpha\in \text{Ord}\}$, i.e.  
$\text{ZF}+\forall \alpha\in \text{C}~~~\text{CH}_{\alpha}\nvdash \forall \alpha\in \text{Ord}~~~\text{CH}_{\alpha}$ 
(b) The assumption $\{\text{CH}_{\alpha}~|~\alpha\in \text{C}\}$ is sufficient to prove $\text{AC}$, i.e.
$\text{ZF}+\forall \alpha\in \text{C}~~~\text{CH}_{\alpha}\vdash \text{AC}$
Question (2): If the answer of the question (1) is positive, can we choose $\text{C}$ to be a set not a proper class?
Question (3): What are the minimal classes (by inclusion order) like $\text{C}$ in the question (1)?
 A: All you need for AC in the standard argument from GCH is that the GCH holds for an unbounded class of cardinals. The reason is that this is sufficient to conclude that any set of sets of ordinals is well-orderable, and this is sufficient to imply AC. 
So the answer to question 1 is yes; any unbounded class $C$ suffices. 
Meanwhile, merely knowing the GCH holds for cardinals below some cardinal is insufficient, since one can build the analogue of the symmetric models for $\neg\text{AC}$ above any cardinal, while preserving GCH below. Thus, there also can be no minimal $C$, since every sufficient $C$ is unbounded, and we may omit any proper initial segment of it and still have a sufficient $C$. 
A: Let me add on Joel's answer and point out that in fact in $\sf ZF$ the following weakening of $\sf GCH$ holds:

For every $A$, if $A$ is well-orderable, then $\mathcal P(A)$ is well-orderable $\implies$ The Axiom of Choice.

From the above it is immediate that if there is a proper class of $\alpha$ such that $2^{\aleph_\alpha}=\aleph_{\alpha+1}$, then the axiom of choice holds. The proof is due to Herman Rubin.
One should note, however, that without the axiom of choice $\sf GCH$ can often be taken as "For every infinite set $A$, there is no $X$ such that $|A|<|X|<|\mathcal P(A)|$". This too implies the axiom of choice, and therefore the statement $\forall\alpha(2^{\aleph_\alpha}=\aleph_{\alpha+1})$. 
The proofs that I know of the implications are different in nature. When assuming the weaker principle which is only for $\aleph$ numbers (or its weakening mentioned above), the proof usually goes by transfinite induction to show that every $V_\alpha$ is well-ordered and conclude the axiom of choice. When assuming that there is no intermediate cardinal between an infinite set and its power set, the proof usually goes to show that every set can be injected into its Hartogs number and therefore can be well-ordered.
