Being a T1-Space Is Equivalent to All Finite Subsets Being Closed
Theorem
A necessary and sufficient condition for $X$ to be a $T_{1}$-space is that every singleton set $\left\{ x \right\}$ of $X$ is closed in $X$.
Proof
$(\Rightarrow)$
For a $T_{1}$-space $X$, let $x \in X$, $x' \in X \setminus \left\{ x \right\}$, so that $x \ne x '$. Since $X$ is a $T_{1}$-space, there exists an open set $U_{x’} \subset X$ such that $x' \in U_{x’}$ and $x \notin U_{x’}$. Summarizing, $$ x' \in U_{x’} \subset X \setminus \left\{ x \right\} $$ and $$ X \setminus \left\{ x \right\} = \bigcup_{x’ \in X \setminus \left\{ x \right\} } U_{x’} $$ is an open set. Therefore the singleton set $\left\{ x \right\}$ is closed in $X$.
$(\Leftarrow)$
Since every singleton set of $X$ is closed in $X$, for $x_{1} \ne x_{2}$ the sets $\left\{ x_{1} \right\}$, $\left\{ x_{2} \right\}$ are closed in $X$. Then $$ U_{1} := X \setminus \left\{ x_{1} \right\} \\ U_{2} := X \setminus \left\{ x_{2} \right\} $$ are open in $X$. On the other hand, since $$ x_{2} \in U_{1} x_{1} \in U_{2} $$ $X$ is a $T_{1}$-space.
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Explanation
Since the union of closed sets is still a closed set, it is fine to say that this is equivalent to all finite subsets being closed.