Radii of Neighborhoods with non-Real Distance Metric Here is the definition of a neighborhood in Walter Rudin's Principles of Mathematical Analysis:

A neighborhood of $p$ is a set $N_r (p)$ consisting of all $q$ such that $d(p, q) < r$, for some $r>0$. The number $r$ is called the radius of $N_r (p)$

Now, it is not explicitly stated that $r \in \mathbb{R}$, however in exercise 10 of Chapter 2, he states asks about the compactness of an infinite metric space whose distance function is defined as

$d(p, q) = 1$ if $p \neq q$
$d(p, q) = 0$ if $p = q$

One's response to the compactness as well as closed-ness and openness of subsets of this space seems then to depend on whether $r$ can be any $\mathbb{R}$ between 0 and 1, or whether it is confined to be a value in the set of possible distances - but how would it even make sense to place $r$ as a distance which cannot exist? It seems like overlaying a metric function whose image is indeed $\mathbb{R}$ on top of the discrete one, but maybe I am too insistent on the geometry.
I suspect he does mean any real number, since any neighborhood is open, and any finite set is closed.
Additionally, to prove the following theorem by contradiction, he uses some notion of finite, yet open neighborhoods, which seems impossible to create. How can he do this?

If E is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
Proof If no point of $K$ were a limit point of $E$, then each $q \in K$ would have a neighborhood $V_q$ which contains at most one point of $E$ (namely $q$, if $q \in E$). It is clear that no finite subcollection of $\{V_q\}$ can cover $E$ and the same is true of $K$, since $E \subset K$. This contradicts the compactness of $K$.

 A: Your first comment:

Now, it is not explicitly stated that $r \in \mathbb{R}$,

True, it is not explicitly stated, however since it is stated that $r>0$, and $>$ is a relation defined on $\mathbb R$, it is implicitly stated that $r\in\mathbb R$.

Second comment:

One's response to the compactness as well as closed-ness and openness of subsets of this space seems then to depend on whether $r$ can be any $\mathbb{R}$ between 0 and 1, or whether it is confined to be a value in the set of possible distances

True, if you were allowed to change the range from which $r$ can be selected, then the closedness of sets would change. But, alas, you are not allowed to change the range. When speaking of neighborhoods, $r$ can always be any positive real number. That's the definition of neighborhoods. If you restrict $r$, then you aren't making a mistake, really, but what you are then talking about are no longer neighborhoods. You can call those things whatever else you want (for all I care, call them fluffy cushions), but the word neighborhood is taken.

Third comment:

but how would it even make sense to place $r$ as a distance which cannot exist?

Why not? We can perfectly well define the set $\{x\in X: d(x_0, x) < 10\}$ even though the maximum distance we can reach is $1$. Sure, the set will be the same as the set $${x\in X: d(x_0, x) < 100}$, but why do you think this is somehow wrong? I can perfectly well speak of "the set of all humans that are shorter than 15 meters", and sure, this is a redundant way of speaking (since this set is the same as the set of all humans), but it is not wrong. It's a perfectly well defined set.

Fourth:

and any finite set is closed.

This is false in general. It is possible to define a topology in which finite sets are not closed. However, it is true in metric spaces.

Fifth:

Additionally, to prove the following theorem by contradiction, he uses some contrived notion of finite, yet open neighborhoods.

Calling an argument "contrived" is not a mathematical argument, so I cannot really comment on that. The proof you cite is mathematically rigorous and correct, and you calling it names doesn't change the correctness of the underlying mathematics.
In fact, your comment shows a clear misunderstanding of the proof, since Rudin himself does not speak of finite neighborhoods anywhere in the proof. Each set $V_q$ is a neighborhood, and depending on the metric it can be finite, but in the case of a Euclidean metric on $\mathbb R^n$, the set $V_q$ is infinite.
A: 
but how would it even make sense to place r as a distance which cannot exist?

There is no issue here. Even in the metric you've given, the so called discrete metric, you can ask about $N_{1/2}(p)$. We have $N_{1/2}(p) = \{x\in X: d(p,x)<1/2\}=\{p\}$ since all other points $x\neq p$ in $X$ will give $d(p,x)=1>1/2$.

It seems to be a weak point, not a feature, of this mathematical genus that we can invent impossible, ontologically unsound things to prove a point, such as finite neighborhoods.

Calm down, there is nothing wrong with what Rudin has written, topology can be counter intuitive but it's not ontologically unsound. A neighborhood can consist of a single point, as we have just shown above. Think about the definition of limit point and what it would mean to NOT be a limit point. This should help you understand why there would have to be a neighborhood of each point with at most one point from $E$ (Because if such a neighborhood does not exist then ____).
