When is the closure of an intersection equal to the intersection of closures? We know $\overline{\bigcap A_{\alpha}}\subseteq\bigcap\overline{A}_{\alpha} $, but when is the reverse inclusion true?  Can you give some properties of the underlying space that would guarantee this?
 A: This extends the answers of  hardmath and Brian M. Scott, but completely answers the question.
Spaces satisfying the (seemingly) weaker condition that $\overline{ A \cap B } = \overline{A} \cap \overline{B}$ for all $A , B \subseteq X$ are discrete.

If $A \subseteq X$ is a non-closed set, pick $x \in \overline{A} \setminus A$.  Note, then, that $$x \in \overline{ \{ x \} } \cap \overline{A} = \overline{ \{ x \} \cap A } = \overline{ \varnothing } = \varnothing,$$ which is absurd!  Therefore all subsets are closed.

As hardmath noted, all discrete spaces satisfy the stronger condition in the OP, and so we have an equivalence of all three notions.

Given a topological space $X$, the following are equivalent:  
  
  
*
  
*$\overline{ \bigcap_{i \in I} A_i } = \bigcap_{i \in I} \overline{A_i}$ for all families $\{ A_i \}_{i \in I}$ of subsets of $X$.
  
*$\overline{ A \cap B } = \overline{A} \cap \overline{B}$ for all $A , B \subseteq X$.
  
*$X$ is discrete.
  

A: It's trivially so if the underlying space has a discrete topology.
In such cases $\overline{A} = A$ for any subset $A$ of the space.
A: Inspired by this answer, which showed that $A^\circ\cup B^\circ$ iff $\partial A\cap\partial B\subseteq\partial(A\cup B)$, here $A^\circ$ and $\partial A$ denoting respectively the interior and the boundary of $A$, I put a similar property on the two sets $A,B$ guaranteeing $\bar A\cap\bar B=\overline{A\cap B}$.
Noting that $\overline{A^c}=A^{c\circ}$ (abbreviation for $(A^c)^\circ$) and $\partial A=\partial(A^c)$, we have
$$\begin{aligned}\overline{A\cap B}=\bar A\cap\bar B
 &\Leftrightarrow(A\cap B)^{c\circ}=A^{c\circ}\cup B^{c\circ}\\
 &\Leftrightarrow(A^c\cup B^c)^\circ=A^{c\circ}\cup B^{c\circ}\\
 &\Leftrightarrow\partial A^c\cap\partial B^c\subseteq\partial(A^c\cup B^c)\\
 &\Leftrightarrow\partial A\cap\partial B\subseteq\partial(A\cap B).
\end{aligned}$$
This is a necessary and sufficient condition for $\bar A\cap\bar B=\overline{A\cap B}$. As to the case of a family of sets, I don't know any similar condition.
A: $\newcommand{\cl}{\operatorname{cl}}$There is a very large class of spaces in which it fails. 
Let $X$ be any space with a non-isolated point $p$ such that the intersection of all nbhds of $p$ is $\{p\}$. Let $\mathscr{N}$ be the set of nbhds of $p$, and for each $N\in\mathscr{N}$ let $N'=N\setminus\{p\}$. Then
$$\bigcap_{N\in\mathscr{N}}\cl N'\supseteq\{p\}\ne\varnothing=\cl\bigcap_{N\in\mathscr{N}}N'\;.$$
This class includes all non-discrete $T_1$ spaces. (In case it isn’t immediately obvious that $p\in\cl N'$ for each $N\in\mathscr{N}$, note that $\mathscr{N}$ is a filter, and $\{p\}\notin\mathscr{N}$, so $N'\cap M\ne\varnothing$ for each $N,M\in\mathscr{N}$.)
