Lemma 4.1.4. Let $X$ be a topological space, then $X$ satisfies (T1) if and only if $\bracs{x}$ is closed for all $x \in X$.
Proof. $(\Rightarrow)$: Let $x \in X$, then for each $y \in X \setminus{x}$, there exists $U_{y} \subset X$ open such that $x \not\in U_{y}$. Thus $U^{c} = \bigcup_{y \in X \setminus \bracs{x}}U_{y}$ is open.
$(\Leftarrow)$: Let $x, y \in X$ with $x \ne y$, then $y \in \bracs{x}^{c}$, $x \not\in \bracs{x}^{c}$, and $\bracs{x}^{c}$ is open.$\square$