Correctness of proof that every neighborhood is an open set. Rudin makes the following definitions:
(a) A neighborhood of p is a set $N_r(p)$ consisting of all $q$ such that $d(p, q) < r$, for some $r > 0$.
(b) $E$ is open if every point of $E$ is an interior point of $E$.
(c) $p$ is an interior point of $E$ if there exists a neighborhood $N_r(p)$ of $p$ that is a subset of $E$. 
This is my attempt at the proof that every neighborhood $N = N_r(p)$ is an open set:
Let $x \in N$. Then there exists a neighborhood of $x$ that is also a subset of $N$, namely $N$ itself. Since $x$ and $N$ were arbitrary, every neighborhood is an open set.
Rudin's proof involves the use of the metric, and I wonder why it wouldn't use a more general approach, or if my proof is at all correct.
 A: Given Rudin's definition of "neighborhood," if $x\in N_r(p)$ but $x\not=p$, then $N_r(p)$ is not, itself, a neighborhood of $x$.  Strictly speaking, a set that is a neighborhood of a point is a ball with that point as its center, so $N_r(p)$ can only be a neighborhood of $p$.  What Rudin therefore needs to prove -- and where metric considerations enter in -- is that if $x\in N_r(p)$, then there is some ball centered at $x$, $N_{r'}(x)$, that is completely contained in $N_r(p)$.
A: In his Principles of Mathematical Analysis, Rudin defines "neighborhoods" as what we commonly call "open balls", and denoted $B(x,r)$ (Rudin also adopts this terminology in his Real and Complex Analysis). The standard definition of a neighborhood $N$ of $x$ is a set with an open subset $V$ such that
$$
x\in V \subset N.
$$
A more restrictive definition would be that $N$ is an open set containing $x$. These two definitions almost always serve the same purpose.
As for your proof, you need to show that $B(x,r)$ contains open balls about other points in it. When you say "$x$ and $N$ are arbitrary", it doesn't mean other arbitrary $N$s fall in your current $B(x,r)$.
A: The problem with your proof is that you're assuming that $N$ is a "neighborhood" of $x$ for every $x\in N$. That is not actually the case with the (slightly non-standard) definitions you quote.
For example, on the real line, $N=(1,5)=N_2(3)$ is a "neighborhood" of $3$, and $2\in N$, but $(1,5)$ is not a "neighborhood" of $2$. So you cannot, given these definition, use $N$ itself as a witness that $2$ is an interior point of $N$.
A: You can only define the neighborhood of a point or a subset so the fact you want to show does not make sense. N is a neighborhood of p if there exists an open subset which contains p and which is contained in N.
