Are there weak versions of the axiom of choice equivalent to weak versions of Zorn's lemma and similar principles? I recalled reading about other weaker forms of $AC$, for example countable choice, where we could make choices from a sequence $(S_{k})_{k \in \mathbb{N}}$ of non-empty sets. I also recalled mention of dependent choice, whereby given a relationship $R$ on a set $X$ such that $(\forall x \in X)( \exists y \in X)(x R y)$, one can construct a sequence $(x_{k})_{k \in \mathbb{N}} \in X^{\mathbb{N}}$ such that $x_{k} R x_{k + 1}$ for all $k$.
My question is, is there much use in defining notions of choice similar to countable choice for larger infinite cardinals? Foe example, if $\kappa$ is an infinite cardinal, then is there a use in defining $AC_{\kappa}$ by saying that for any family $(S_{i})_{i \in I}$ of non-empty sets, where $\mathrm{card}(I) \leq \kappa$? Or is there some analogous way to define dependent choice for something "bigger" than sequences? Moreover, I am familiar with a long list of results which are equivalent to $AC$, e.g. the existence of a well-order for every set, Zorn's lemma, Tychonoff's theorem, etc.; for these weaker choices, are there analogous assertions which are equivalent to this weaker choice, e.g. is there some weak form of Zorns's lemma equivalent to $AC_{\kappa}$?
I'm interested in "how much" of the body of results typically associated with choice you can recover from these weaker forms of it. For example, "how much" choice do I need to prove the existence of a unique closure of a field of cardinality $\kappa$? What about to show that for a set $X$ of infinite cardinality $\kappa$, the Kleene closure $X^{*} : = \bigcup_{n = 1}^{\infty} X^{n}$ of $X$ has cardinality $\kappa$? In other words, $AC$ has a lot of results that say things like, "For every set $X$ with this and that property, you can do this with $X$", or "For any family of non-empty sets $(X_{i})_{i \in I}$ where all $X_{i}$ have some property, then $\prod_{i \in I} X_{i}$ has some property". But suppose I have my infinite set $X$, or my infinite set $I$, and want to make those same statements. Then how little choice could I use and still get away with those statements? For example, if I have a set $I$, and a family $(X_{i})_{i \in I}$ of sets $X_{i}$, where each $X_{i}$ is equitotient with $I$, then what would it take to show that $\bigcup_{i \in I} X_{i}$ is equitotient with $I$ (analogous to the countable union of countable sets being countable)?
Thanks.
 A: The question is what do you mean by "use". You can show that countable choice does not imply the existence of a non-measurable set; but strengthening it to $\sf DC_{\aleph_1}$ does.
Countable choice is not used at all in the proof that a product of compact Hausdorff spaces is compact. This is in fact equivalent to the Ultrafilter Lemma, which is consistent even when countable choice fails spectacularly (in Cohen's first model the Ultrafilter Lemma is true, but there is a subset of $\Bbb R$ which is dense but has no countably infinite subset).
Let me add a quick equivalent for Zorn's lemma. I wrote $\sf DC_{\aleph_1}$ above. Dependent Choice is in fact the restriction of Zorn's lemma in the following way.

Let $\kappa$ be a well-ordered infinite cardinal. $\sf DC_\kappa$ is [equivalent to] the statement that every partial order $(P,\leq)$ where every chain of order type $\alpha<\kappa$ has an upper bound, has either a maximal element or a chain of order type $\kappa$.

(The original formulation of $\sf DC_\kappa$, due to Azriel Levy, is exactly saying that arbitrary transfinite recursion can be carried up to $\kappa$.)
We can show that $\sf DC_\kappa$ implies $\sf AC_\kappa$, and that every cardinality is comparable with $\kappa$; but neither of those imply $\sf DC_\kappa$ (not even their conjunction! and the implications between these two is a bit intricate). So Dependent Choice principles are really the stronger principles. Most of the time when a laymathematician says "countable choice" what they really mean is $\sf DC_{\aleph_0}$ which is the correct amount of choice to allow recursive definitions.
To your edit let me first point out that if by $\kappa$ you mean an $\aleph$ cardinal, then for the man things you don't need choice at all. If $X$ has cardinality $\kappa$, then by definition it can be well-ordered, and for ordinals we can canonically enumerate the finite sequences and well-order them again, so if $X$ is infinite and well-orderable, $X^*$ is also well-orderable, and they have the same cardinality.
For the specific case of algebraic closures, we can prove there is always an algebraic closure for a well-orderable field; to prove it is unique we can use the Ultrafilter Lemma, which means that there is no reasonable restriction of $\sf AC$ (a la $\sf AC_\kappa$) which is equivalent to it for reasons I wrote above.
Generally, the uses of the axiom of choice are quite ostensible. They might seem similar, and sometimes you can replace one by another, but they end up being very different (again, countable choice vs. ultrafilters). To see what sort of choice you can recover from an assumption, or how much choice is really needed for a proof, you need to work carefully and see what sort of structure you used in a given proof, and what sort of properties and "choice" you've used there.
For example, consider the downwards Lowenheim-Skolem theorem. Given an infinite structure in a countable language, it has a countable elementary submodel. Close investigation of the proof will show that you use dependent choice to produce a sequence of approximations to the submodel, and a careful observation will show that from this theorem you can in fact prove Dependent Choice itself. 
But now replace countable by $\leq\kappa$ for some $\aleph$ number, you might expect this to turn up as $\sf DC_\kappa$. The truth, however, is that the recursive constructions are of length $\omega$, so you only use $\sf DC$. But you still need more, and the missing part is exactly $\sf AC_\kappa$. So the downwards L-S theorem for size $\kappa$ is equivalent to the conjunction of $\sf DC+AC_\kappa$. And again, you can prove the other direction and establish an equivalence.
There is no general method, though. There's no general algorithm for solving these sort of problems, and they often end up having clever and less-expected solutions with results that you may not guess in advance.
Let me just finish by pointing out that $\sf ZF$ proves that $\Bbb R$ is always uncountable. What it does not prove, which perhaps is what your professor alluded to, is that $\Bbb R$ is not the countable union of countable sets (which can, in the absence of choice, be uncountable).
