A Question about the Intuition Behind the Definition of an Interior Point on Baby Rudin On page 32 of Rudin's Principles of Mathematical Analysis, in section 2.18 (e), the following definition is introduced:
Let $X$ be a metric space and let $p \in X$ and $E \subseteq X$. p is called an interior point of $E$ if there is a neighbourhood $N$ of $p$ such that $N \subseteq E$.
As I understand it, the idea of an interior point intuitively is that if $E$ is a subset of a metric space, then a point which is "inside" $E$, or in other words, "not on the edge" of $E$, we would call an interior point.  
If that's really the idea behind the mathematical term - an interior point, then I think that the following example is an interior point according to the definition, but not really an interior point according to the idea:
the set $X = \{1, 2, 3\}$ is a metric space, since $\mathbb{R}^1$, just like any Euclidean space is a metric space and any subset of a metric space is also a metric space. If you put $E = X$, and $p = 3$, then $p$ is an interior point of $E$, since you can take $E = N_3(p)$ as a neighborhood of $p$ and obviously $E \subseteq E$. But intuitevly, you would say that $p$ is on the edge of $E$ and not inside of $E$. 
So does the example I've given is not actually an interior point, or am I'm missing the idea behind the term interior point?      
 A: The concept being defined has three objects given: the metric space $X$; the point $p \in X$; and the subset $E \subset X$. 
The author uses the terminology "$x$ is an interior point of $E$", without referring to $X$, because $X$ is considered the "background" topological space and does not need special mention. 
Nonetheless, it might be somewhat clearer to incorporate $X$ into the terminology, and some authors do just this. So, copying that somewhat lengthier but clearer terminology, we say that $x$ is an interior point of $E$ relative to $X$  if there exists a subset $N \subset X$ such that $N$ is a neighborhood of $p$ and $N \subset E$.
In your example, it is true that if $X = E = \{1,2,3\}$ and if $p = 3$ then $p$ is an interior point of $E$ relative to $X$, using the neighborhood $N = \{3\}$. The point is that $X$ is a discrete metric space and therefore every subset is open, including the subset $N = \{3\}$. That's just how things go in a discrete metric space. It's a lonely place, you are the only one in your neighborhood.
However, if $X = \mathbb R$ and $E = \{1,2,3\}$ and $p=3$ then $x$ is not an interior point of $E$ relative to $X$.
A: The subset $X = \{1, 2, 3\}$ as a subset of $\mathbb{R}^1$ with the Euclidean Metric has no interior points. As any neighborhood around 1, 2, or 3 would contain points outside of $X$. Where your analysis breaks down is that the requirement is that $E \subseteq X$ in order for the point to be an interior point, not $E \subseteq E$
A: If $p$ is an interior point of a subset $E$, then there is some neighbourhood $N \subset E$ which contains $p$. This means that $p$ is contained within an open ball $B_\varepsilon(p) \subset N \subset E$.
So if $p$ is an interior point of a set $E$, there is some positive distance from $p$ to the complement of $E$. In that sense, it is "inside the set".
Your example with $X = \{ 1,2,3\}$ does not really work, even if you see it as a subspace of $\mathbb{R}$ with the subspace topology (which was my original interpretation). In that topology, every singleton set $\{x\}$ of $X$ is open, since $\{ x \} = X \cap B_{1/2}(x)$. Hence, every subset of $X$ is open, so every point of every subset is an interior point.
