If a T1 topological space is closed under countable intersections, does this necessarily make the topology discrete? It is easy to construct a counterexample if the topological space is not assumed to be T1, and it is also easy to see that the claim is true in a metrizable topological space. If the claim is false under the assumption that the space is T1, then what is the weakest condition required to ensure the topology is discrete?

Here I am gathering information in the comments along with some information of my own to form an answer.

A $P$-space is a regular space where every countable intersection of open sets is open. There are many examples of $P$-spaces which are not discrete. For example, the co-countable topology is a topological space $(X,\mathcal{T})$ where $X$ is an uncountable set and $\mathcal{T}$ consists of all subsets $R\subseteq X$ where $|R^{c}|\leq\aleph_{0}$ or $R=\emptyset$. The co-countable topology is a $T_{1}$-space where the intersection of countably many open sets is still open.

An Alexandroff space is a space where the intersection of arbitrarily many open sets is open. The only $T_{1}$-Alexandroff spaces are the discrete spaces.

Here are a few examples of benign topological properties for which the only $P$-spaces which satisfy those properties are the discrete spaces. From this point on, let us limit our scope to only the regular spaces.

**Connectedness:** Every $P$-space is zero-dimensional (a zero-dimensional space is a space with a basis of clopen sets). Therefore, any connected or locally connected regular $P$-space consists of a single point. Furthermore, any regular space $X$ with a connected subset consisting of more than one element cannot be a $P$-space.

**Disconnectedness:** A topological space $X$ is said to be extremally disconnected if for each open set $U$, the closure $\overline{U}$ is still open. A cardinal $\lambda$ is said to be below the first uncountable measurable cardinal if there does not exist a non-principal ultrafilter $M$ on $\lambda$ such that the intersection of countably many elements of $M$ still belongs to $M$. Every uncountably measurable cardinal is very large and it is consistent with ZFC that there are no measurable cardinals. Every extremally disconnected $P$-space of non-measurable cardinality must be discrete, but there are extremally disconneced $P$-spaces of measurable cardinality such as the one in this answer (this space happens to be the poset used for Prikry forcing).

**Compactness:** Every compact $P$-space is discrete and finite. In fact, in every $P$-space, every countable subspace must be closed. However, every infinite compact space must have infinite discrete non-closed subspace, so infinite compact spaces cannot be $P$-spaces at the same time. If $X$ is a $P$-space which is pseudocompact, countable compact, or locally compact, then $X$ must be discrete.

**Countability:** Suppose that $L$ is one of the following properties: metrizable spaces, second countable, first countable, points are $G_{\delta}$, sequential spaces. Then the only $P$-spaces that satisfy property $L$ are the discrete spaces.

**Non-examples:** There are many topological properties for which there are many non-discrete $P$-spaces which satisfy those properties including the following (somewhat redundant) list: completely regular, normal, paracompact, Lindelof, realcompact, scattered, the Baire property and its generalizations, the negation of the Baire property.

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