Let $(X,\tau)$ a no discrete topological space. If necessary for an affirmative answer consider a metric space $(X, d )$ or a Banach space $(X, \|\,\cdot\, \|)$. In these cases, the topology $\tau $ would be generated $d$ or $ \|\,\cdot\, \|$ respectively.
It is well known examples of countable family of open sets whose intersection is not a open set. For example $\bigcap_{n\in \mathbb{N}}(-1-\frac{1}{n},1+\frac{1}{n})=[-1,1]$.
Question. There is a suficente condition about a family $\{ \mathcal{O}_\alpha\}_{\alpha\in I}$ ( $I=\mathbb{N}\mbox{ or } \mathbb{R}$ ) of open sets in $\tau$ topology that ensures that the intersection is also an open set which is not discrete in $\tau$ topology?
Furthermore, every point $x$ of a metric space $X$ is the intersection of a countable family of balls with radius $\frac{1}{n}$ centered at $x$. If $x$ is not a isolated point then the intersection $\bigcap_{n\in \mathbb{N}}B(x,\frac{1}{n})=\{x\}$ is closed. But if $x$ is an isolated point the intersection $\bigcap_{n\in \mathbb{N}}B(x,\frac{1}{n})=\{x\}$ is an open set. But this sufficient condition is trivial.
An interesting example is the intersection $\bigcap_{t\in I}A_t$ of a family $\{A_t\}_{t\in I}$ ( $I=[0,1])$ of open sets $A_t$ living in the function space $C^{1}([0,1],\mathbb{R})$.