Can infinite $T_0$ space have less open sets than points? It is pretty easy to check, that if $X$ is $T_1$, then it has to have at least as many open sets as points, because every singleton would be closed. On the other hand, if we don't assume anything about $X$, we could choose indiscrete topology and get positive answer.
In the middle, we have $T_0$. My intuition says, that it shouldn't be possible, but I don't know why.
 A: The proposed by you solution for $T_1$ case works almost the same in $T_0$ case, except you should not focus on singletons, but rather on closures of singletons.
Let $(X,\tau)$ be a $T_0$ topological space. Note that two points $x,y$ are topologically indistinguishable if and only if $\overline{\{x\}}=\overline{\{y\}}$. Thus in $T_0$ spaces $x=y$ if and only if $\overline{\{x\}}=\overline{\{y\}}$. This shows that
$$f:X\to \tau$$
$$f(x)=X\backslash\overline{\{x\}}$$
is an injection, regardless of whether $X$ is infinite or not.
A: Remark: This is correct but unnecessarily complicated.
No, it cannot. Suppose $X$ is an infinite set and $(X,\mathcal{T})$ is a $T_0$ topological space. Note that $\mathcal{T}$ must be infinite or otherwise this space would have at least some pairs of distinct points that are topologically indistinguishable.
Given any infinite set $A$, let $S(A)$ denote the set of unordered pairs $\{a,b\}$ of distinct points in $A$. Since $A$ is infinite, we have $|A\times A|=|A|$. If $B=A\times A\backslash \{(a,a)\mid a\in A\}$, then there is a canonical surjection $B\to S(A)$ taking an ordered pair $(a,b)$ of distinct points to the unordered pair $\{a,b\}$. One can then extend this to a surjection $A\times A\to S(A)$ to get $|S(A)|\leq |A|$. It's not hard to construct an injection $A\to S(A)$ and so we have $|A|=|S(A)|$.
Applying the previous paragraph to our situation, we have $|X|=|S(X)|$ and $|\mathcal{T}|=|S(\mathcal{T})|$. Thus, to prove that $|X|\leq |\mathcal{T}|$ it suffices to produce an injection $S(X)\to S(\mathcal{T})$.
Define $f:S(X)\to S(\mathcal{T})$ by $f(\{x,y\})=\{X\backslash\overline{x},X\backslash\overline{y}\}$ where $\overline{x}$ denotes the closure of the single-point set $\{x\}$. We'll check some things about $f$ using the following key fact about $T_0$ spaces.
Fact: Since $X$ is $T_0$, if we are given $x,y\in X$, then we have $x=y$ if and only if $\overline{x}=\overline{y}$.
Claim 1: $f$ is well-defined.
Pf. If $x\neq y$, then because $X$ is $T_0$, $\overline{x}\neq \overline{y}$. Thus $X\backslash\overline{x}\neq X\backslash\overline{y}$ showing the pair $\{X\backslash\overline{x},X\backslash\overline{y}\}$ is an element of $S(\mathcal{T})$. We conclude that $f$ is well-defined.
Claim 2: $f$ is injective.
Pf. Suppose $a\neq b$, $x\neq y$ and $f(\{a,b\})=f(\{x,y\})$. Then $\{X\backslash\overline{a},X\backslash\overline{b}\}=\{X\backslash\overline{x},X\backslash\overline{y}\}$ are equal 2-element sets. If $X\backslash\overline{a}=X\backslash\overline{x}$ and $X\backslash\overline{b}=X\backslash\overline{y}$, then $\overline{a}=\overline{x}$ and $\overline{b}=\overline{y}$, giving $a=x$ and $b=y$. If $X\backslash\overline{a}=X\backslash\overline{y}$ and $X\backslash\overline{b}=X\backslash\overline{x}$, the same reasoning gives $a=y$ and $b=x$. Either way, $\{a,b\}=\{x,y\}$. This proves $f$ is injective.
