Your proof is nearly perfect, just notation errors.
There are similar ideas as your previous question: Proving the supremum of a set in the general case
Existence: After fixing some $y>0$, your set $S$ should be $S=\{x\in\mathbb{R}:x^3<y\}$, you do not need $x\ge0$. But your ideas are the same. Using your set $S$:
For $\beta^3<x$, its more clear to write 'there exists $z\in\mathbb{R}$ s.t. $\beta^3<z^3<x$' instead of $\beta<z$ and $z^3<x$, and likewise for the other case.
For $\beta^3>x$, $z$ is an upper bound of $S$! The contradiction should be 'So $\sup S\neq\beta$.'
Using my set $S$, you should conclude $\beta^3=y$.
Uniqueness: See Ross Millikan's comment.
[Q: we have considered this set as a basis for our proof. However, doesn't this mean that the proof only applies to that set? Why did we chose that set? Wouldn't this set need to have a special property that all integers share so that we can prove the theorem for all integers?]
Note that $S$ is defined by $a$ and $b$ which are any integers such that $b>0$. Since this proof never gives exact values for $a$ and $b$, any integers can be put in their place (thus proving for all integers).
[Q: Do we separate both parts of the proof from each other? In other words, can we apply the results we got from the first part of the part to the second one? It doesn't really make sense since we can't have 0∈S and 0∉S at the same time.]
This is an example of cases in a proof. Think of it like an argument that has two possible starting places. You are trying to prove something for all $S$, but sometimes its hard to do that so you have to break it into parts. You first prove that its true if $0 \in S$, and then you prove that its true if $0 \not \in S$. The parts are not really connected, and don't reference each other, but since every $S$ must have $0$ or not have $0$, the statement must be true for every $S$ since its true for both cases.
[Q: is this the q we need to prove that it exists? Didn't we now just assume that it exists?]
Due to the Well Ordering Principle there must be a smallest member we call $r$, and since $r \in S$ it must be able to follow the "format" of $S$ which is $a−bk$. What we are doing here is refer to $k$ as $q$ since we know this will be the value we are looking for.
[Q: Didn't we assume that 0∉S? That would mean that r>0? Did we set that because r=0 in the above example? However, both parts should be treated separately.]
You are right that $r > 0$, but that also means that $r \geq 0$. All the proof is trying to show is that $r \geq 0$, so it doesn't need to keep the strict condition that $r > 0$.
[Q: Is this because r=a−bq in general or because we have set r=a−bq in the above part of proof?]
We have set $r=a−bq$, so $a-bq = r \geq b$ or $a-bq \geq b$.
[Q: why?]
Since $b$ divides $r' - r$ it must be true that $r' - r$ is a multiple of $b$. But, $r' - r < b$ and $r' - r \geq 0$, so it can only be a multiple of $b$ if $r' - r = 0$.
Best Answer
I think your difficulty is just a point of logic. When we want to prove uniqueness of a thing, it's a standard technique to assume there are two different such things and then show that they are not different after all.
Suppose I needed to prove that there is a unique integer between 3 and 5. I would assume that there were two such integers, $x$ and $y$ such that $3<x<5$ and $3<y<5$. Then I would try to argue that $x=y$, and probably that argument would be "by contradiction." That is, I would assume $x \neq y$ and try to get an absurdity.
Since one of $x$ and $y$ is bigger than the other (because we're assuming they're not equal) and since it makes no difference which one is bigger, we might as well assume (WLOG, as they say) that $x>y$. One way to say this is $x-y$ is a natural number (because natural numbers are positive.)
Now that things are set up nicely, I can get on with my (silly) proof: Since $3<x<5,$ we have $3-y < x-y <5-y$. And since $y>3$, $5-y <5-3 =2$, so $5-y\leq 1.$ Combining these inequalities gives us $x-y < 5-y \leq 1$, so $x-y$ is a natural number less than one, absurd. I conclude that $x-y$.