Proof check for "every neighbourhood is an open set" I have been goind through Rudin for the past month and I have arrived at this theorem in the topology chapter. Now I understand Rudin's proof but I was trying to come up with my own and I am not sure if it is correct or not.
Theorem: Every neighbourhood is an open set.
Proof: Suppose that there exists a neighbourhood $N$ of $p$ that is not open. Then there exists a point $q$ in $N$ that is not an interior point of $N$. Then any neighbourhood $N_q$ of $q$ is not included in $N$. Choose $N_q$ such that $p$ is in $N_q$. Since $N_q$ is not in $N$, then $p$ is not in in $N$. Contradiction.
I have attached an image showing what I had in mind when I wrote this.
The gray edge boundary means that it's not included

I think your proof is incorrect.  There is a problem with your statement: "Since $N_q$ is not in $N$, then $p$ is not in $N$".
It is true that $N_q$ is not a subset of $N$, but it is certainly possible that some elements (points) of $N_q$ are also elements of $N$.
