Suppose $F$ is a subset of a topological space $X$. Prove that $F$ is closed if every point $x \in X$ has an open neighborhood $V$ s.t. $F \cap V$ is a closed subset of $V$.
Proof:
Let $x \in X$. Then, by hypothesis, $\exists$ open hood' $V \subset X$ such that $F \cap V$ is closed in $V$.
Since $F \cap V$ is a closed subset of $V$, $\exists$ closed hood' $W \subset X$ such that $W \cap V = F \cap V$.
We then have $W = F$, and thus $F$ is closed.
EDIT: Of course I errored and it is not true that $W \cap V = F \cap V \rightarrow W = F$. Can somebody help me fix this proof?
Take an arbitrary $x\in X \setminus F$. By hypothesis, there is an open neighborhood $V$ of $x$ such that $F\cap V$ is closed in $V$. Since $x\notin F$, then $x\notin F\cap V$. So there is an open neighborhood $U$ of $x$ such that $(U\cap V) \cap (F \cap V) = \emptyset$, or $U \cap V \cap F = \emptyset$. As $U\cap V$ is an open neighborhood of $x$ disjoint from $F$, it follows that $X\setminus F$ is open, i.e., $F$ is closed.