In my analysis course, we are considering $(\mathbb{R},+,\cdot,\leq)$ as axiomatically constructed ordered field. Now, together with that, we added a completness axiom stated as follows:
Axiom: Let $(I_n)$ be a sequence of intervals in $\mathbb{R}$ such that for all $n\in\mathbb{N}$ $I_{n+1}\subset I_n$. Also let $\lim_{n\to\infty} |I_n|=0$. Then $$\bigcap_{n=1}^\infty I_n\neq \emptyset$$
Now, I wish to use the least upper bound property (LUBP), that is, any bounded (from above) subset of $\mathbb{R}$ has a least upper bound. My lecturer proposed that this is equivallent to the Axiom of Choice, because we have all subsets of $\mathbb{R}$ (an uncountable set) and for each of these sets $X \in \mathcal{P}(\mathbb{R})$, there is a set of all upper bounds of $X$. Which is also uncountable. So in order to select the $\min{X}$, we need AC, right In this construction, do we need AC to get to LUBP? In addition, what would we need to be able to prove LUBP as a theorem? (Also, if I am wrong, I would love to see a proof that LUBP can be proven from the axiom i gave.)
No, you don't need the axim of choice here. What we are stating here, is that a certain family of sets (namely, the family of all sets of upper bounds of the non-empty upper bounded real numbers) is such that each element of that family has a minimum. Besides, if we define the real field as an ordered field such that the Archimedian property holds and that the axiom that you mentioned holds too, then we can prove that property, and we don't need the axiom of choice to do that.
By the way: let $\mathscr N=\mathcal{P}(\mathbb{N})\setminus\{\emptyset\}$. The set $\mathscr N$ is uncountable and each of its elements has a minimum. Do you need the axiom of choice to prove that?