every single point is closed?

You might find the following useful:

A topological space $X$ is a $T_1$-space if whenever $x,y\in X$ with $x\ne y$, there is an open set $U$ such that $x\notin U$ and $y\in U$.

Proposition. $X$ is $T_1$ iff $\{x\}$ is closed for each $x\in X$.

Proof. Suppose first that $X$ is $T_1$, and let $x\in X$. Then for each $y\in X$ there is an open $U_y\subseteq X$ such that $y\in U_y$ and $x\notin U_y$. Let $U=\bigcup_{y\in X\setminus\{x\}}U_y$; then $U$ is open, and $U=X\setminus\{x\}$, so $\{x\}$ is closed. Now suppose that $x\in X$ and $\{x\}$ is closed. Then $U=X\setminus\{x\}$ is open, and if $x\ne y\in X$, then $U$ is an open set such that $x\notin U$ and $y\in U$. $\dashv$

There are many spaces that are not $T_1$. For example, let $X=\Bbb N$, for each $n\in\Bbb N$ let $$U_n=\{k\in\Bbb N:k<n\}\;,$$ and let $\tau=\{\Bbb N\}\cup\{U_n:n\in\Bbb N\}$; then $\langle X,\tau\rangle$ is a space that isn’t $T_1$. In fact, if $m,n\in X$ with $m,n$, and $U$ is any open set containing $n$, then $m\in U$ as well. Even simpler is the Sierpiński space, whose underlying set is $\{0,1\}$ and whose open sets are $\varnothing,\{1\}$, and $\{0,1\}$: there is no open set containing $0$ but not $1$.

Let $X=\{1,2,3\}$. Let $\mathcal{T} = \{ \emptyset, \{1,2\}, \{1,2,3\} \} $. Then the set $\{ 2 \}$ is not closed. It is not open either.