Show that any finite $T_1$-space $X$ is discrete.
My thoughts:-
Let $X=${$a_1 , a_2, …,a_n$} and $U$ be any subset of $X$.
Let $U=${$a_1 , a_2, …,a_k$} where$k \le n$ thenwe need to show that $U$ is open that is $X \setminus U$ is closed.
Now $X \setminus U =$ { $a_{k+1} , …,a_n$ }$ = ⋃_{(k-1)}^{n}a_i$ now each $a_i$ is closed.
So the result follows. Does my approach correct?
What you’ve done is almost right. You need to consider all subsets of $X$, not just those of the form $\{a_1,\dots,a_k\}$. For instance, if $n=5$ you have not considered the set $\{a_2,a_4\}$.
You can fix the problem in several ways. You could, for instance, let $U$ be any subset of $X$ and observe that
$$X\setminus U=\bigcup_{x\in X\setminus U}\{x\}$$
is a union of finitely many closed sets and is therefore closed, so that $U$ must be open; this is the argument that you’re using now, but done right, so that it applies to all subsets of $X$.
Alternatively, you could leave the complementation to the end. Let $A$ be any subset of $X$. Then
$$A=\bigcup_{x\in A}\{x\}$$
is the union of finitely many closed sets, so $A$ is closed. Thus, every subset of $X$ is closed. Now let $U$ be any subset of $X$; then $X\setminus U$ is closed, so $U$ is open.
Your approach is basically correct, but 1) can be simplifies and 2) contains a mistake.
The mistake is in your assumption that $U=\{a_1,\cdots ,a_k\}$. Such a $U$ is not a generic subset of $X$, so if you manage to prove that it is open/closed it won't prove the result.
The simplification of the argument is: every singleton is a closed set, thus (use finitely ness the space), every singleton is open, thus every arbitrary subset is open.
