Analogue of Urysohn metrization for Lawvere metric spaces? Urysohn proved that any regular, Hausdorff, second-countable space $X$ is metrizable, i.e. there exists a metric space whose underlying topological space is $X$. But what if we ask the same question for Lawvere metric spaces?
Definition: Let $(X,d)$ be a Lawvere metric space. For any $\epsilon>0$ and point $x\in X$, define the $\epsilon$-ball emanating from $x$, denoted $B(x,\epsilon)$, to be the set
$$B(x,\epsilon):=\{x'\in X\mid d(x,x')<\epsilon\}.$$
Define the induced topology on $X$ to be the set of those subsets $U\subseteq X$ with the property that for all $x\in U$ there exists $\epsilon>0$ with $B(x,\epsilon)\subseteq X$.
Example: The Sierpinski space is not Hausdorff, so it is not metrizable. But it is Lawvere metrizable. Indeed, let $S=\{o,c\}$, let $d(o,c)=2$ and $d(c,o)=0$. Then taking $\epsilon=1$ we have $B(o,1)=\{o\}$, so the singleton set $\{o\}$ is open. But for all $\epsilon>0$, we have $B(c,\epsilon)=\{o,c\}$, so $\{c\}$ is not open.
Question: Do you know of a characterization of those topological spaces that are Lawvere metrizable?
 A: According to this SE-post, a Lawvere metric on a set $X$ is a function $d:X\times X\to[0,+\infty)$ satisfying two axioms:
1) $d(x,x)=0$ and
2) $d(x,z)\le d(x,y)+d(y,z)$
for all $x,y,z\in X$.
Then the following theorem can be considered as a counterpart of the Urysohn metrization theorem (I strongly suspect that this theorem was known to Lawvere).

Theorem. Each topological space $X$ with countable base is metrizable by a Lawvere metric.

Proof. Fix a countable base $\{U_n\}_{n\in\omega}$ of the topology of $X$. For every $n\in\omega$ consider the Lawvere metric $f_n:X\times X\to \{0,1\}$ defined by $$f(x,y)=\begin{cases}1&\mbox{if $x\in U_n$ and $y\notin U_n$};\\
0&\mbox{otherwise}.
\end{cases}
$$ 
The Lawvere metrics $f_n$, $n\in\omega$ compose another Lawvere metric $$f=\max_{n\in\omega}\frac1{2^n}f_n$$
which generates the topology of $X$. 
Indeed, for any open set $U\subset X$ and any $x\in U$ we can find $n\in\omega$ with $x\in U_n\subset U$ and conclude that $B(x,\frac1{2^n})\subset U_n\subset U$.
On the other hand, let us show that a set $U\subset X$ is open if for any $x\in U$ there exists $\varepsilon>0$ with $B(x,\varepsilon)\subset U$. Let $\Omega_x:=\{n\in\omega:x\in U_n\}$. Choose $m\in\omega$ with $\frac1{2^m}<\varepsilon$ and consider the open neighborhood $V:=\bigcup\{U_n:n\in\Omega_x,\;n\le m\}$ of $x$.
Observe that for every $y\in V$ and every $n\le m$ we have $f_n(x,y)=0$.
So, $d(x,y)<\frac1{2^m}<\varepsilon$ and $x\in V\subset B(x,\varepsilon)\subset U$, which means that $x$ is an interior point of $U$ and $U$ is open.
Remark. The proof essentially uses the Lawvere metrizability of the Sierpinski two-point space and the fact that each $T_0$-space with countable base embeds into the countable product of the Sierpinski two-point spaces.
