Alexandroff space is T1 if and only if it is discrete

Proposition. Let $X$ be an Alexandroff space. Then $X$ is ${\mathrm{T}}_1$ if and only if $X$ is discrete.

Proof. ,,$\Leftarrow$” It is easy to see, that every discrete space is Alexandroff and ${\mathrm{T}}_1$.

,,$\Rightarrow$” Recall that topological space is ${\mathrm{T}}_1$ if and only if every subset is equal to the intersection of all its open neighbourhoods. So let $x\in X$. Then the intersection of all open neighbourhoods ${\{x\}}^o$ of $x$ is equal to $\{x\}$. But since $X$ is Alexandroff, then ${\{x\}}^o=\{x\}$ is open and thus points are open. Therefore $X$ is discrete. $\mathrm{\square }$