Why does the $T_1$ Axiom imply that a point is a closed set?

Question

According to my reading the $T_1$ axiom say that given a topological space $X$ and $\forall x, y \in X$ We have:

$ \exists U, V \subset X: [(x \in U \land y \notin U) \land (x \notin V \land y \in V)].$

Here, $U, V$ are open sets, and they're not assumed to be disjoint.

Why does this imply that $\forall x \in X$, the set $\{ x \}$ is closed?

Answer 1

For every $y\ne x$ choose an open subset $V_y$ such that $y\in V_y,x\notin V_y$. Then $$X\setminus\{x\}=\bigcup_{x\ne y\in X}V_y$$ is open as it is the union of open sets, hence $\{x\}$ is closed.

Answer 2

Let $x \in X$. We'll show that $\{x\}$ is closed: let $y \neq x$ be arbitrary, then there is an open set $O_y$ such that $y \in O_y$ and $x \notin O_y$. This means that $O_y \cap \{x\} = \emptyset$, so that $y$ is not an adherence point (closure point) of $\{x\}$. So no point except $x$ itself lies in the closure of $\{x\}$, and the singleton is closed.