Let $(X,\mathscr T)$ be topological space. Prove that $ (X,\mathscr T)$ is $T_0$-space iff for each pair of $a$ and $b$ distinct members of X, $\overline{\{a\}}\neq \overline{\{b\}}.$
My attempt:-
Let $(X,\mathscr T)$ be topological space. Prove that $ (X,\mathscr T)$ is $T_0$-space. $a\neq b \implies$ there is an open set $U$ contains $a$ but not $b$. $\{b\}\subseteq X\setminus U \implies \overline{\{b\}}\subseteq X\setminus U.$ $a\notin X\setminus U$. Any closed set containing $a$ must be completeltely disjoint from $X\setminus U.$ So, $\overline{\{a\}}\neq \overline{\{b\}}.$
Conversaly, $\overline{\{a\}}\neq \overline{\{b\}}.$ Consider $X\setminus \overline{\{b\}}$, which is an open set. $X\setminus \overline{\{b\}} \subseteq X\setminus \{b\}$. $a\in X\setminus \{b\}$ . $b\notin X\setminus \{b\}\implies b\notin X\setminus \overline{\{b\}}$. How do I complete the proof?
The first part of your attempt starts with a wrong statement.
You say that if $X$ is $T_0$ and $a \ne b$, then there exists an open $U$ containing $a$ but not $b$. However, this is the definition of $T_1$. In a $T_0$ space you can only say there exists an open $U$ containing exactly one of $a, b$. Of course you can say that without loss of generality we may assume $a \in U$ and $b \notin U$. To see the difference, consider the Sierpinski space $\Sigma = \{ 0, 1 \}$ with topology $\mathfrak{T} = \{ \emptyset, \{ 1 \}, \Sigma \}$. It is $T_0$ because the only two distinct points are $0, 1$ and $\{ 1 \}$ is an open set containing exactly on of these points whereas there does not exist an open set containing $0$ but not $1$.
Moreover, you cannot conclude that any closed set containing $a$ must be completely disjoint from $X \setminus U$. But it is irrelevant since $a \notin \overline{\{b \}}$, hence $\overline{\{a \}} \ne \overline{\{b \}}$. As an example consider $\Sigma$ and take $a = 1, b = 0, U = \{1 \}$. You have $\overline{\{ a \}} = \Sigma$ and $\overline{\{ b \}} = \{ 0\}$. In fact $\overline{\{ b \}} \subset X \setminus U$ and $a \notin X \setminus U$, but $\overline{\{ a \}}$ is not completely disjoint from $X \setminus U$.
For the converse, let $a, b$ distinct points of $X$. Then $\overline{\{a \}} \ne \overline{\{b \}}$. The set $S = \overline{\{a \}} \cap \overline{\{b \}}$ is closed.
It is impossible that both $a, b \in S$ because otherwise $\overline{\{a \}} \subset S \subset \overline{\{b \}}$ and $\overline{\{b \}} \subset S \subset \overline{\{a \}}$, i.e. $\overline{\{a \}} = \overline{\{b \}}$.
Case 1. $S$ contains none of $a, b$. Then $b \notin \overline{\{a \}}$ (because otherwise $b \in \overline{\{a \}} \cap \overline{\{b \}} = S$), hence $b \in X \setminus \overline{\{a \}}$ and we are done since $a \in \overline{\{a \}}$, i.e. $a \notin X \setminus \overline{\{a \}}$.
Case 2. $S$ contains exactly one of $a, b$, w.l.o.g. $a$. Then $a \notin X \setminus S$ and $b \in X \setminus S$ and we are done again.