Could you help me to prove this proposition below?
$X$ is a finite set and $(X,\tau)$ is a topological space. Show that if every subset which have single element is closed set then $\tau$ describes discrete topology.
I started with let $X=\{x_1,x_2,\cdots,x_n\}$ . If every subset with single element is closed set then for $1\leq i \leq n$ then $\{x_i\}$ is closed, at the same time $X \setminus \{x_i\}$ is open set.
for$\quad$ $1\leq i \leq n$ $\quad$ $X \setminus \{x_i\}=\{x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n\}$ and we know that $(X,\tau)$ is a topological space so it provides the axioms. It is the point where i stuck. How can i build discrete topology with using the set $\quad$ $X \setminus \{x_i\}=\{x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n\}$ by using finite intersection and (finite or infinite) union set operations. thank you in advance :)