Say $X$ is a $T_1$ topological space. Let's show that $\{x\}$ is closed for every $x\in X$ :

The $T_1$ axiom gives us, for every $y$ distinct from $x$ , an open $U_y$ that contains $y$ but not $x$ . Since we're in a topological space, we can take the union of all these open sets to get a new open set, $$U=\bigcup_{y\neq x} U_y $$

$\{x\}$ is the complement of $U$ , closed because $U$ is open: None of the $U_y$ contain $x$ , so $U$ doesn't contain $x$ . But any $y\neq x$ is in $U$ , since $y\in U_y\subset U$ . That takes care of that.

Now let's say we have a topological space $X$ in which $\{x\}$ is closed for every $x\in X$ . We'd like to show that $T_1$ holds:

Given $x\neq y$ , we want to find an open set that contains $x$ but not $y$ . $\{y\}$ is closed by hypothesis, so its complement is open, and our search is over.