**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.