Proof

By hypothesis, $(X,\mathcal O)$ is a topological space.

Assume, $X$ is a $T_1$ space.
Let $\{x\}\subseteq X$ be a singleton subset of $X.$
Case 1: The set difference is empty, i.e. $X\setminus \{x\}=\emptyset.$
- Then $\{x\}=X$ and $\{x\}$ is closed.
Case 2: $X\setminus \{x\}\neq \emptyset.$
- Then there is an $y\in X\setminus \{x\}$ with $y\neq x$
- Since $X$ is a $T_1$ space, there is an open set $U\in\mathcal O$ with $x\not\in U$ and $y\in U\subseteq X\setminus \{x\}.$
- Since this is the case for every such $y$, the set $X\setminus\{x\}$ is open.
- Thus, $\{x\}$ is closed.

Assume, every singleton subset of $X$ is closed.
Let $x,y\in X$ with $x\neq y.$
Since $\{x\}$ and $\{y\}$ are closed, $X\setminus\{x\}$ and $X\setminus\{y\}$ are open.
Moreover, $X\setminus\{x\}$ is an open set not containing $y$ and $X\setminus\{y\}$ is an open set not containing $x.$
Thus, $X$ is a $T_1$ space.
∎

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