Proof

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

Assume, $X$ is a $T_2$ space.
Let $B$ be a [filter base]https://www.bookofproofs.org/branches/base-of-a-filter/.
Further, assume that $x,y$ are two distinct limit points of $B.$
- Since $X$ is a $T_2$ space, there are disjoint open sets $O_x,O_y\in\mathcal O$ containing $x$ and $y$ respectively, i.e. $x\in O_x$, $y\in O_y$ and $O_x\cap O_y=\emptyset.$
- By the definition of a limit point, $O_x$ contains at least one point of $B$ distinct from $x$ and $O_y$ contains at least one point of $B$ distinct from $y$.
- But since $B$ is a filter base, any two sets of $B$ contain a set in $B.$
- Therefore, $O_x\cap O_y$ contains a set in $B.$
- But this is a contradiction to $O_x\cap O_y=\emptyset.$
It follows that $x=y,$ i.e. $B$ has only one unique limit point.

Assume, $B$ is a [filter base]https://www.bookofproofs.org/branches/base-of-a-filter/ $B$ in $X$ and has only one unique limit point $x.$
Further assume that $X$ is not a $T_2$ space.
- By this assumption, there is a points $y\in X$ distinct from $x$ such that all open sets containing $x$ and $y$ respectively are not disjoint i.e. $x\neq y$ but $O_x\cap O_y\neq\emptyset$ for all $O_x$ with $x\in O_x$ and all $O_y$ with $y\in O_y.$
- Since $B$ is a filter base, every intersection $O_x\cap O_y$ contains a set in $B.$
- But every such intersection has to contain a point $y$ distinct from $x.$
- This is a contradiction to $x\neq y.$
It follows that $X$ is a $T_2$ space.
∎

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