Topological spaces which are not pseudometrizable. Let $(X,\tau)$ be a topological space. Then we know some conditions under which $(X,\tau)$ is metrizable (see for example this and this). It is also clear from these theorems that not every topological space is metrizable.
However, I am wondering whether the same is true for pseudometric spaces also. To be more specific, 

Does there exist topological spaces which are not pseudometrazible? 

Where we agree to call a topological space $(X,\tau)$ to be pseudometrizable iff  there exists a pseudometric $d$ on $X$ such that the topology induced by the psuedometric is $\tau$. 
 A: A pseudometric space is symmetric (also called $R_0$): if $x \in \overline{\{y\}}$ then $y \in \overline{\{x\}}$ (basically because $d(x,y)=0$ implies $d(y,x)=0$ too, also in pseudometric spaces). 
Sierpinski space ($X=\{0,1\}$ with topology $\{\emptyset,\{0\},X\}$) is not symmetric so not pseudometrisable. (Because $1 \in \overline{\{0\}}$ but not the other way around). This is in a way the simplest example, certainly the smallest one.
If $X$ is $T_1$ then $X$ is metrisable iff $X$ is pseudometrisable. (the $T_1$ ensures that $X$ is also $R_0$ and so the non-existence of any points $x,y$ with $x \neq y$ but $d(x,y)=0$. So the pseudometric for $X$ on the right is then a metric.)
So spaces like the cofinite topology on $\mathbb{N}$ is not pseudometrisable, as it's not metrisable (not Hausdorff to start with...)
Also, the Sorgenfrey line, the Michael line, double arrow space etc etc.
A: The Sierpinski space is not pseudometrizable. The Kolmogorov quotient of a pseudometric space is metric. However the Sierpinski space is already $T_0$, but it's not Hausdorff, and thus not metric.
The Sierpinski space is the topological space on two points with the topology
$$\{\varnothing, \{0\}, \{0,1\}\}.$$
