Without axiom of choice, why can't we choose a random element for each set in an infinite collection of sets

Question

If we have an infinite collection of sets, where each set contains two elements, 0 and 1, without using axioms of choice, we can define a choice function that will always pick the element 0 from each set. However, it seems that if we only assume the set contains two elements, without specifying them in detail, we can not show the existence of a choice function without using the axiom of choice.

Since all the sets has a cardinality of two, we can define bijections between the two elements in each set and {0,1}. So regardless of how the bijection is defined, we can always choose the element that maps to 0.

I don't understand the difference between this operation compares to always choosing 0 in each set, but one requires the axiom of choice, while the other doesn't.

I have read many answers on stack exchange about axioms of choice. Like this one. And I saw answers that used the word "uniform" selection. I'm really confused about what "uniform" means in this case, and no questions seem to answer my confusion.

EDIT: The comment section is getting too long. So I posted a new question.

Answer 1

Since all the sets has a cardinality of two, we can define bijections between the two elements in each set and {0,1}.

This actually does not help at all, although it's a little subtle to see why: it's true that for each such set, such a bijection exists, but in order to write down a choice function you need to choose a collection of such bijections for each set simultaneously. This means you need a choice function for the collection of bijections! And there are two for each set! So you're in exactly as bad a position as you were before.

To say this more explicitly, suppose $I$ is an index set and $X_i, i \in I$ is a collection of $2$-element sets indexed by $I$. You are saying: well, since by hypothesis each $X_i$ has cardinality $2$, for each $i$ there exists a bijection $X_i \cong \{ 0, 1 \}$. And yes, that's true. But to write down the choice function you want using these bijections, you need to choose one such bijection for each $i \in I$, and there are two. In other words, you need a choice function for the collection of sets $Y_i = \text{Iso}(X_i, \{ 0, 1 \})$, each of which also has two elements!

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Coming at this from another direction, here is a more-or-less explicit example of a collection of sets of cardinality two where I challenge you to write down an actual choice function: consider the collection of all fields of characteristic $\neq 2$ in which $-1$ has a square root (if you want to cut this down to a set then take isomorphism classes and place some restriction on the cardinality). There are always exactly two such square roots, one of which is the negative of the other, which in the complex numbers are $i$ and $-i$. Can you tell me how to choose, for each such field, one of the two square roots of $-1$? Note that there are "surprising" examples of such fields, such as the field of $p$-adic numbers $\mathbb{Q}_p$ for $p \equiv 1 \bmod 4$ a prime.

Another example in a similar vein: consider the collection of all connected orientable manifolds. Any such manifold has exactly two orientations, one of which is the negative of the other. Can you tell me how to choose, for each such manifold, one of its two orientations?