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 ($n=3$ for instance)

Edit :Bof was quicker than me

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)