My own proof of "Every neighborhood is open" (Rudin's definition) Previously, someone posted his/her proof of "every neighborhood is open" here. I am focusing on the same statement, but it seems that my approach is different from his/hers. Could someone comment on my own proof? Here we go:
Let $X$ be a metric space and $p\in X$. Fix $r>0$. Define a neighborhood around $p$ as $N_r (p)=\{q\in X| d(p, q)<r\}$. Now, take $w = \frac{r}{2}$. Then, $N_w (p)\subset N_r(p)$. In other words, every point of $N_r(p)$ is interior. Therefore, $N_r (p)$ is open. Since $r$ is arbitrary, this proves every neighborhood is open.
Update:
Under my definition of "neighborhood," can I prove in this way instead: For all $m\in N_r(p)$, take $w=\frac{r−d(m,p)}{2}$. Then, $N_w(m)\subset N_r(p)$. 
Update2
From the hints given in comments, I solve it:
Assume $m\in N_r(p)$ and $q\in N_w(m)$. Take $w=r-d(m, p)$. Then, $d(p, q)\leq d(p, m) + d(m, q)< d(p, m) + r-d(m,p) =r$. 
Thanks to @JoséCarlosSantos and @fleablood !
 A: In the first place, you can call that a neighbourhood, if you wish, but that's not the standard definition of neighbourhood. So, the only thing that you might have proved was that those sets to which you call neighbourhoods are open sets.
In the second place, your proof is wrong. All that it proves is that there is an open ball centered at p contained in your set. It does not prove that for every point of $N_r(p)$ there is such a ball.
A: First of all, you shouldn't say "define a neighborhood".
That sounds as though you are defining the term to be something different than it is.  Rudin uses this "open ball" definition of "neighborhood" so, although, other comments and answers object, it is okay to DECLARE $N_r(p)$ to be a neighborhood.
Now you have shown that if $N_{\frac r2}(p)\subset N_r(p)$ so $p$ is an interior point.
Then you say "In other words, every point of Nr(p) is interior."
Um, say what?  Since when has $p$ been every point? $N_r(p)$ contains points that are not $p$.  You must prove these points are also interior points.
The only real way that I can see is to chose an $s$ so that $N_s(q) \subset N_r(p)$.  (Note: $N_s(q)$ and $N_r(p)$ have different centers.).  And if you know that $d(p,q) < r$ then there is an $s$ that ca be chosen so that $N_s(q) \subset N_r(p)$.  Do you see how to do that?
