Smallest neighbourhood characterization in finitely generated spaces To prove that finitely generated spaces are corecompact, I used the following characterisation of finitely generated topological spaces:

In a finitely generated topological space every point has a smallest
  neighbourhood.

Now I was wondering how to prove this characterisation. The definition I use for finitely generated topological spaces is the following:

$\forall A \in \mathcal{P}(X), \forall x \in X: x \in cl(A) \Rightarrow \exists a \in A, x \in cl\{a\}$

I tried proving it by contradiction: Assume that there is no such thing as a smallest neighbourhood. Suppose that $\forall V \in \mathcal{V}(x) \exists W \in \mathcal{V}(x); W \subseteq V$.
Can someone give me a hint so I can come to a contradiction? Or is there a nicer way to prove this?
As always, any hints or comments will be appreciated!
 A: Let $U$ be the intersection of all neighborhoods of $x$. We show that $U$ is open. Let $y$ be in the closure of $X\backslash U$. We have to show that $y\in X\backslash U$. By assumption, there is $z\in X\backslash U$ such that $y$ is in the closure of $\{z\}$. Now by construction of $U$, there is an open neighborhood $V$ of $x$ such that $z\notin V$. Since $V$ is open, the closure of $\{z\}$ is not in $V$ and hence $y\notin V$. Hence $y\notin U$ and $y\in X\backslash U$. 
A: If you show that, in a finitely generated space, any intersection of open sets is open, then it is easy to see that every point $x$ has a smallest neighborhood - it is the intersection of all open sets containing $x$. 
In fact, the claim that intersection of any system of open sets is open is a characterization of finitely-generated spaces, see Wikipedia.
However, let us have a look only at the direction important for you: You want to show that any finitely-generated space (according to your definition) has this property. This is equivalent to showing that arbitrary union of closed sets is closed.
Let $F=\bigcup_{i\in I}F_i$. Let $x\in\overline{F}$, then there is an $a\in F$ such that $x\in\overline{\{a\}}$. This implies the existence of an $i\in I$ such that $x\in\overline F_i=F_i$ (we take $i\in I$ for which $a\in F_i$). Clearly, this implies $x\in F$. We have shown that $F$ is closed.
