Graph associated to a topological space Let $(X,\tau)$ be a topological space. We say that $x\neq y\in X$ are close if every neighborhood of $x$ intersects every neighborhood of $y$. We associate to $(X,\tau)$ a graph $G(X,\tau)=(V,E)$ with $V = X$ and $$E=\big\{\{x,y\}: x\neq y \in X \text{ and } x,y \text{ are close}\big\}.$$
Given an infinite simple graph $G=(V,E)$, is there a topology $\tau$ on $V$ such that $G(V,\tau) = G$?
 A: Counterexample. Let $V$ be an infinite set. Choose four distinct points $a,b,c,d\in V.$ Let $G$ be the graph on the vertex set $V$ in which every pair of distinct points is an edge except $\{a,b\}$ and $\{c,d\}.$ Assume for a contradiction that $\tau$ is a topology on $V$ such that $G(V,\tau)=G.$
Since $a$ and $b$ are not close they have disjoint open neighborhoods, call them $A$ and $B.$ Then $b$ is close to no point of $A.$ Since $b$ is close to every point except $a,$ it follows that $A=\{a\},$ i.e., $\{a\}$ is an open set. A similar argument shows that $\{c\}$ is an open set. But then $a$ and $c$ are not close, although they are joined by an edge.
More generally, let $G=(V,E)$ be any graph. If $G=G(V,\tau)$ for some topology $\tau,$ then the following condition must hold:

If $a,b,a',b'$ are four distinct vertices such that $ab\in E$ and $aa',bb'\notin E,$ then there is a vertex $v\notin\{a',b'\}$ such that $va'\notin E$ and $vb'\notin E.$

(Clearly, there are many infinite graphs $G$ which do not satisfy this condition.)
Proof. Since $aa',bb'\notin E,$ there are open sets $A,A',B,B'$ such that $a\in A,a'\in A',A\cap A'=\emptyset,b\in B,b'\in B',B\cap B'=\emptyset.$ Since $ab\in E,$ we have $A\cap B\ne\emptyset;$ choose a vertex $v\in A\cap B.$ Since $A,A'$ are disjoint open sets with $v\in A$ and $a'\in A',$ we have $v\ne a'$ and $va'\notin E.$ Similarly, $v\ne b'$ and $vb'\notin E.$ Thus we have $v\notin\{a',b'\}$ and $va',vb'\notin E,$ Q.E.D.
A: I think the answer is (sadly) no : take a complete graph $K_n = (V, E)$ and remove an edge, say between $x$ and $y$. Then assume you have such a topology $\tau$. Take any $U, W\in \tau$ such that $x\in U$, $y \in W$ and $U\cap W=\emptyset$ (two such open sets exist,otherwise $x$ would be close to $y$). Since any point in $U$, different from $x$ is close to $y$, if there were any, say $z$, we would have that $U$ is a neighbourhood of $z$, and $W$ a neighbourhood of $y$, so that $U\cap W \neq \emptyset$. That's a contradiction, so $U=\{x\}$. For sufficiently large $n$, this is obviously impossible ($n=4$ for instance)
Edit :Bof was quicker than me
2nd edit : there was a mistake in my argument : the last part where I say it's impossible : well it's not. There needs to be at least 4 points so that you can remove 2 edges with no common endpoint, apply the same reasoning to another $a$, so that both $\{a\}$ and $\{x\}$ are open sets, and $a,x$ are connected by an edge, which then brings us a contradiction. Sorry about that. (However the answer is still no)
