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.