Understanding analysis/topology proofs 
Let $X$ be a finite metric space with $\{p\} = E \subset X.$ Does $E$ have interior points?

Proof's given below. But I want to put it in my own words so that I understand it.

Text below is how I see it. Not very pretty, but as long as it makes sense, it's all good to me.
$N_r(p)$ is a neighborhood of $p$ of radius $r.$
Let $r < $ smallest $q \in X.$ Then $N_r(p)$ contains all points of $X$ that are to the left of the smallest $q$ on the number line meaning all $q \in X$ fall outside $N_r(p).$ For example, if the smallest $q = 1$ and $r = 0.5,$ all points $\ge 1 \ \in X$  are too large to fit in the distance between p and $0.5.$ But the interval within the radius $r$ contains $p$ and probably some other points smaller than $q.$ Since we are only worried about $p$ and $q,$ we ignore all other points $< q$ (that are not $p$). Thus $N_r(p) = \{p\} \subset E$ and so $p$ is interior point of $E.$
Does my interpretation make sense?
edit: Also, I just realized real numbers don't have "thickness" to them so they (however large they are)  fit between $p$ and $0.5$ 
 A: I think there are some points you aren't completely understanding here.
First of all, there is no number line involved here, $X$ is an arbitrary set, with some distance function defined on it, specifically, $X$ is not necessarily linearly ordered, so something like "the smallest $q$" doesn't really make sense in this context. 
Secondly, if we do assume $X$ is a subset of $\mathbb R$, or linearly ordered, then $N_r(p)$ contains all the points whose distance from p is less than the distance of the nearest $q\in X$ to $p$ (which is probably what you meant).
What we are trying to prove is that $N_r(p)$ contains no points besides $p$, meaning that there aren't any points closer to $p$ than $q$. So if 

the interval within the radius $r$ contains $p$ and some other points smaller than $q$

those points would be inside $N_r(p)$, and then $N_r(p)$ wouldn't be a neighborhood of $p$ contained in $E$.
Basically, we don't ignore other points in $X$ because we only care about $q$, we ignore them because we specifically chose $q$ such that there aren't any other points between them.
A: The statement is indeed meaningless, except in ordered sets.
If $X = \{q_1, \ldots,q_n\}$, and $p \in X$, then there are the distances (which are real numbers, so have an order) $d(p,q_1),\ldots,d(p,q_n)$. As $p$ is one of these $q_i$, exactly one of these numbers is $0$, the others are $>0$.
So $r = \frac{1}{2}\min\{d(p, q_i) :q_i \neq p\}$ is well defined (and what the author should have written) and $r>0$ (as the minimum is one of the numbers $d(p, q_i) > 0$.
The claim is that $N_r(p) = \{p\}$. Of course $p \in N_r(p)$ always, and if $x \neq p$ then $x = q_i$ for some $q_i$. We took the minimum of all $d(p,q_i)$ like that so $\min\{d(p, q_i): q_i \neq p\} \le d(p,q_i)$ so in particular $r < d(p,q_i)$ But this means that $q_i \notin N_r(p)$. So no point from $X$ except $p$ can lie in $N_r(p)$, so $N_r(p)=\{p\}$ is a witnessing ball that shows that $x$ is not a limit point of $X$.
