I have been reading Munkres' Topology and in Ch. 1 section 10, the Well-Ordering Theorem is introduced as equivalent to AC, there is even a proof of it outlined in the supplementary exercises. However, the main use of this theorem in the book seems to be to show that uncountable well-ordered sets exist. From a bit of research (and even in Munkres' book) you can show that there exist uncountable well-ordered sets without AC.
To my actual question, it is stated all over in various sources that the well-ordering of $\mathbb{R}$, or $\mathbb{R}$ being impossible to well-order is independent of ZF. I can't seem to find a proof or a link to a paper that shows that this is in fact the case. I don't know tons of set theory, just what is given in the first chapter of Munkres' book, so maybe I couldn't even understand the proof, but I would at least like to know it's out there and try to understand it. This is the closest thing I can find to talking about the subject but still doesn't mention the reason why the well-ordering of $\mathbb{R}$ is independent of ZF. Any help is much appreciated.
The axiom of choice implies that every set can be well-ordered. In particular, it implies that the real numbers can be well-ordered.
So a proof that the axiom of choice is consistent with $\sf ZF$ would be also a proof that the fact that the real numbers can be well-ordered consistently with $\sf ZF$.
The relative consistency of the axiom of choice is normally proved via Gödel's constructible universe, but you need to have a lot more set theory under your belt to fully understand the proof. You can find it in most advanced set theory books (e.g. Jech, or Kunen).
The main difficulty is to prove that no well-ordering of the reals can be explicitly defined, or that a well-ordering of the reals does not imply the full axiom of choice. For these you need even more set theory than before, as the proofs include forcing and symmetric extensions. You can find these proofs also in Jech or Kunen.