Prove that every $T_1$-space is $T_0$-space

Question

Exercise: A topological space is said to be $T_0$-space if for each pair of distinct points $a,b$ in $X$, either there exists an open set containing $a$ and not $b$, or there exists an open set containing $b$ and not $a$.A topological space $(X,\tau)$ is said to be $T_1$-space if every singleton set $\{x\}$ is closed in $(X,\tau)$

Prove that every $T_1$-space is $T_0$-space.

I attempted the following proof:

If we have two topological spaces $T_1=(X,\tau_1)$ and $T_0=(X,\tau_0)$. If $a,b\in X$, by the definition of $T_1$-space: $B=X\setminus{a}$ is an open set such that $a\notin B $ and $b\in B$. In analogous way $A=X\setminus{b}$ is an open set such that $b\notin A $ and $a\in A$. This proves that the open sets of $\tau_0 $are the open sets of $\tau_1$, then $\tau_0\subset\tau_1$.

Question:

Is the proof right? Can it be said "open sets of $\tau_0 $are the open sets of $\tau_1$, then $\tau_0\subset\tau_0$"?

Answer 1

Say that $(X,\tau) $ is $T_1$ space.

$\iff$ $\{x\}$ is a closed subset of $X$ for all $x\in X$

$\iff$ $\forall y\in \{x\}^c $ $ \exists U_y \in \tau: y\in U_y\subset\{x\}^c $ for all $x\in X$

$\Rightarrow$ For all $y\neq x$ in $X$ there is an open subset $U$ of $X$ such that $y\in U$, $x\notin U$

$\Rightarrow$ $(X,\tau) $ is $T_0$ space.

Answer 2

If every singleton is closed, then given $a \neq b \in X$, the set $X \, \backslash \, \{ a \}$ is an open subset of $X$ containing $b$ and not $a$.