Neighborhoods of points in cofinite topological spaces

My topology text defines the "neighborhood" of a point as follows:Let p be a point in a topological space X .A subset N of X is a neighborhood of p iff N is a superset of an open set G containing p. Now I am stuck on the following problem:Let X be a cofinite topological space .Show that every neighborhood of a point p (element of x) is an open set .Now since in cofinite topology the closed sets are finite , so this should be true by definition , that is aclosed set could not contain an open set ...I just cannot see what we have to prove here ........

Suppose $N$ is a neighborhood of $p$ and that $G$ is an open set such that $p\in G$ and $G\subseteq N$.
Then $X\setminus G\supseteq X\setminus N$, so $X\setminus N$ is finite. Hence $N$ is open.
• Are we showing here that there exists one more larger open set $N$ in the topology, such that $G$ is a subset of it, and $G$ contains $x$ , hence making $N$ a nbd of $x$ as well as member of the topology ? I am a bit confused as to why $N$ should be open – The Doctor Apr 20 '18 at 6:23
• @TheDoctor I'm proving that, if $N$ is a neighborhood of $p$, then $N$ is open. Since it is a neighborhood, there exists an open set $G$ with $p\in G\subseteq N$ (by definition of neighborhood). Now we check that $X\setminus N$ is finite, so that $N$ turns out to be open as well. – egreg Apr 20 '18 at 6:26
• because $N'$ is finite, so $N$ is infinite and hence open. Am I right? – The Doctor Apr 20 '18 at 7:20
• @TheDoctor Not really: $N'=X\setminus N$ finite ends the argument (the fact that $N$ is infinite is irrelevant and indeed false if $X$ is finite to begin with). – egreg Apr 20 '18 at 7:24