Let $(X,\tau)$ be a topological space and $V\subseteq X$. Then is equivalent:
1) $x\in\partial V$
2) For every neighborhood $U$ of $x$ is $V\cap U\neq\emptyset\neq (X\setminus V)\cap U$
3) It exists nets $(x_\lambda)_\lambda$ in $V$ and $(y_\lambda)_\lambda$ in $X\setminus V$ with $x_\lambda, y_\lambda\to x$
My thoughts:
Edit: Following the hint of Daniel Schepler:
$x\in\partial V\Leftrightarrow x\in\overline{V}\setminus V^\circ\Leftrightarrow x\in\overline{V}\wedge x\notin V^\circ\Leftrightarrow x\in\overline{V}\wedge x\notin V\setminus \overline{(X\setminus V)}\\\Leftrightarrow x\in\overline{V}\wedge x\in\overline{(X\setminus V)}$
Since $x\in\overline{V}$ I can use what was shown here: Topological space, equivalent statements, net, closure, neighborhood
So we just have to show the "2nd half".
Now I want to show, that $(X\setminus V)\cap U\neq\emptyset$ Suppose $(X\setminus V)\cap U=\emptyset$, then is $U\subseteq (X\setminus V)^c=V$, also $U\subseteq V$. Then is $U_x\subseteq V^\circ$, hence $x\in V^\circ$. But it is $x\in\partial V=\overline{V}\setminus V^\circ$. Contradiction.
Can someone give a hint for $(1)\Leftrightarrow (3)$
Hint: show that $x \in \partial V$ if and only if $x \in \overline V$ and $x \in \overline{X \setminus V}$. Then the equivalence $(1) \Leftrightarrow (2)$ reduces to the similar standard equivalence regarding when $x \in \overline V$, and similarly for the equivalence $(1) \Leftrightarrow (3)$.