Do we need Axiom of Choice to make infinite choices from a set? According to the answers to this question, we do not need choice to pick from a finite product of nonempty sets, even if each of the sets is infinite. The axiom of choice is required to ensure that a infinite product of nonempty sets is non-empty. i.e. $\prod_{i \in I} A_i \neq 0$.
Now, let $A_i = \mathbb{R}$. The answers to this question (and the one linked above) says we do not need choice to pick an element $x_0 \in \mathbb{R}$. Suppose, I want an arbitrary sequence of real numbers $X = (x_n)_{n =1}^{\infty}$. Then, I will have to make an infinite number of "picks" from $\mathbb{R}$.
Is it right to say that the resulting sequence $X \in \prod_{i \in I} \mathbb{R}$ and that we need choice to ensure that it exists? Why or why not?

You need to go to the rational numbers, in order to ensure that every equation of the form $nx=m$ has a solution with $m,n$ natural numbers. Does that mean that there are no solutions, to any choice of $n,m$ in $\Bbb N$?
No, it doesn't. $4x=8$ still has a solution in $\Bbb N$.
Similarly with the axiom of choice. It is needed to ensure that every product of non-empty sets is non-empty. It doesn't mean that it is needed for proving each and every product of non-empty sets is non-empty, though.
If all the sets you choose from are the same, then constant choices are choices that you can always make. If all the sets you choose from have a particular structure on all of them (e.g. they are all finite sets of real numbers), then you can choose from them all (e.g. they are all finite sets of real numbers, take the minimal element).
The axiom of choice is being overused, and in many cases it isn't needed for a particular argument of interest. Not the say that it is never needed, or rarely needed. It just gets overused plenty. And that might be dangerous (see the third panel).
