$X$ is $T_1$ if and only if every finite subset of $X$ is closed Hi I am supposed to show that $X \ is \ T_{1} \iff$ Any finite subset of $X$ is closed, I tried to show both directions, comments would be appreciated.
Let $X$ be $T_{1}$, and let $A$ be a finite subset of $X$. Now to show that $A$ is closed it suffices to show that $X - A$ is open.
Since $X$ is $T_{1}$ $\forall x \neq y \ \exists  $ an open set $U$ s.t. $x \in U, y \notin U$
for $X - A$ to be open in $T_{1}$ we must have $\forall x \neq y$
$\exists x \in X-A $ s.t. $y \notin X-A$ which is true since $X$ is $T_{1}$ Hence $A$ is closed
For the converse statement must show that
Any finite subset of $X$ is closed $\implies$ $X$ is $T_{1}$
Let $A$ be a finite subset of $X$. Since $X-A$ is open $\exists$ a NB around each point in $X-A$. Let $U$ be such a NB for $x \in X-A$. Since A is finite and assuming non empty, $\exists$ elements $y \in A$ s.t. $y \neq x$ which is not in $U$ $\square$
 A: I cannot really follow your proof and I decided to give an alternative proof, enabling you to compare.

First observe that any finite subset of $X$ is closed if and only if every singleton-subset of $X$ is closed.
Here $\implies$ is immediate (singletons are finite sets) and $\impliedby$ follows from the fact that a finite union of closed sets is again closed, combined with the fact that a finite set is a finite union of singletons.
Based on this we conclude that it is enough to prove that $X$ is $T_1$ if and only if its singleton-subsets are closed.

$\implies$
Let $a\in X$ so that $\{a\}$ is such a singleton. Then for every $b\neq a$ an open set $U_b$ exists with $a\notin U_b$ and consequently $\{a\}^{\complement}=\bigcup_{b\neq a}U_b$. Then as a union of open sets $\{a\}^{\complement}$ is open, or equivalently $\{a\}$ is closed.
$\impliedby$
Let it be that $a\neq b$. Then $b\in\{a\}^{\complement}$ and $\{a\}^{\complement}$ is an open set that does not contain $a$ as an element.

P.S. Above $\{a\}^{\complement}$ denotes the complement of $\{a\}$.
