Let $X$ be a Kolmogorov space and define on it a specialization order, ie.: for any $x,y \in X$, we have $x \leq y$ if and only if $x \in \overline{ \{y\} }$, where $\overline{ \{y\} }$ is the topological closure of $\{y\}$. Consider a subset $A \subset X$ such that $x \in A$. I need to prove that:
1) $x$ is maximal in $A$ if and only if $\{x\}$ is the intersection of some family of open subsets of $A$.
2) $x$ is minimal in $A$ if and only if $\{x\}$ is a closed subset of $A$.
(2) is the easiest:
Suppose $x$ is minimal. Suppose $y \in \overline{\{x\}}$. This means $y \le x$, but $x$ is minimal so $y = x$. This means that $\{x\}$ is closed. If $\{x\}$ is closed and $y \le x$, then $y \in \overline{\{x\}}$ by definition, but $\{x\} = \overline{\{x\}}$, so $y \in \{x\}$ and $y = x$. Hence $x$ is minimal.
Suppose $\{x\} = \cap_{i \in I} \{O_i: O_i \subset X, O_i \text{ open }\}$. If $x \le y$ this means that $x \in \overline{\{y\}}$. If $y \neq x$, this means that $y \notin O_i$ for some $i$ (why?) and so $y \in X \setminus O_i$ and so $x \in \overline{\{y\}} \subseteq \overline{X \setminus O_i} = X \setminus O_i$, as $O_i$ is open. But then $x$ is not in $O_i$, contrary to assumed, so we cannot have $y \neq x$, and $x$ is maximal.
I will leave the reverse implication to you.