If every subset of $X$ is saturated, then $(X,\tau)$ is a $T_1$ - space In a general topology exercise the following question is asked:

Is it true that if the topological space $(X,\tau)$ is such that every subset is saturated, then $(X,\tau)$ is a $T_1$- space?


My approach to answering this:
If "every subset is saturated", then $\forall A \subseteq X, \exists B_j \in \tau: A = \bigcap \limits_{j \in J}B_j$, where $J$ is some index set.
A topological space is an $T_1$ - space if every singleton $\{x\}$ with $x \in X$ is closed.
So, because $X\setminus\{x\} \subseteq X$, then we have that $\exists B_j \in \tau: X\setminus\{x\} = \bigcap \limits_{j \in J}B_j$.
Because $\tau$ is a topology on $X$, $\bigcap \limits_{j \in J}B_j \in \tau$, and therefore $X\setminus\{x\} \in \tau$. This means that $\forall x \in X$, $\{x\}$ is closed, this meaning that $(X, \tau)$ is a $T_1$ - space.

I think that this proof only applies to cases where $X$ is a finite set right? Because I used the property that if $B_j \in \tau$, then $\bigcap \limits_{j=1}^nB_j \in \tau$. So $\bigcap \limits_{j \in J}B_j \in \tau$ is only necessarily true if this is a finite intersection. How can I show that the statement is true or false for infinite sets where $\bigcap \limits_{j \in J}B_j$ can represent an infinite intersection?
 A: You're right, $\bigcap_{j \in J} B_j$ need not be open. However, your claim is correct.
Let $x\neq y,$ then there exist index sets $I,J$ such that $$\{x\}=\bigcap_{i\in I} U_i\text{ and } \{y\}=\bigcap_{j \in J}V_j$$ for some $U_i,V_j \in \tau.$ Since $x\neq y,$ there exists $i_0 \in I$ and $j_0 \in J$ such that $x \notin V_{j_0}$ and $y \notin U_{i_0}.$ This proves that $X$ is a $T_1$-space.
A: Equality $ X\setminus\{x\} = \bigcap \limits_{j \in J}B_j$ implies that $B_j= X\setminus\{x\}$ or $B_j=X$ for every $j\in J$ because these are the only subsets of $X$ that contain $ X\setminus\{x\}$ as a subset.
However it cannot be that $B_j=X$ for every $j\in J$ because that would lead to the statement that $ X = \bigcap \limits_{j \in J}B_j$ which contradicts that $ X\setminus\{x\} = \bigcap \limits_{j \in J}B_j$.
So we conclude that $B_{j_0}=X\setminus\{x\}$ for some $j_0\in J$.
Here $B_{j_0}$ is open so  $X\setminus\{x\}$ must be open, hence $\{x\}$ must be closed.
A: Let $ x \neq y$ be two points of $X$. As $ \{x\}$ is saturated we can write
$$ \{x\} = \bigcap \mathcal{U}$$ for some family of open sets $\mathcal{U}$. So for some $U$ in that family, $y \notin U$ and clearly $x \in U$. This shows that $X$ is $T_1$.
