$(\Rightarrow):$ $(X,\tau), \ T_0$ uzayı; $ x,y\in X, \ x\neq y$ olsun ve $y\in\overline{\{x\}} \wedge x\in\overline{\{y\}}$ olduğunu varsayalım.
$y\in\overline{\{x\}} \wedge x\in\overline{\{y\}}\Rightarrow \left(\{y\}\subseteq \overline{\{x\}} \wedge \{x\}\subseteq \overline{\{y\}}\right)$
$\hspace{3.8cm}$$\Rightarrow \left(\overline{\{y\}}\subseteq \overline{\overline{\{x\}}}=\overline{\{x\}} \wedge \overline{\{x\}}\subseteq \overline{\overline{\{y\}}}=\overline{\{y\}}\right)$
$\hspace{3.8cm}$$\Rightarrow \overline{\{x\}}=\overline{\{y\}}\ldots (1)$
$\left.\begin{array}{rr} (x,y\in X)(x\neq y) \\ \\ (X,\tau), \ T_0 \text{ uzayı}\end{array}\right\}\Rightarrow \overline{\{x\}}\neq\overline{\{y\}}\ldots (2)$
$(1),(2)\Rightarrow \text{ÇELİŞKİ.}$
$(\Leftarrow):$ $x,y\in X, \ x\neq y$ olsun ve $\overline{\{x\}}=\overline{\{y\}}$ olduğunu varsayalım. Buradaki linkte yer alan karakterizasyondan faydalanacağız.
$\left.\begin{array}{rr}(x,y\in X)(x\neq y)\Rightarrow \left(x\in\{x\}\right)\left(y\in\{y\}\right)\Rightarrow \left(\{x\}\subseteq \overline{\{x\}}\right)\left(\{y\}\subseteq \overline{\{y\}}\right) \\ \\ \overline{\{x\}}=\overline{\{y\}}\end{array} \right\}\Rightarrow$
$\Rightarrow y\in\overline{\{x\}}\wedge x\in\overline{\{y\}}\ldots (1)$
$\left.\begin{array}{rr}(x,y\in X)(x\neq y) \\ \\ \text{Hipotez}\end{array} \right\}\Rightarrow y\notin\overline{\{x\}}\vee x\notin\overline{\{y\}}\ldots (2)$
$(1),(2)\Rightarrow \text{Çelişki.}$