Does every pseudometric give rise to a topology? I have this example in mind:

Let $S= \{a,b,c\}$ and define $d:S \times S \rightarrow \mathbb{R}$ such that $d(x,y) = 0$ for all $x,y \in S$.

This is a pseudometric because it is symmetric, non-negative and $x = y \implies d(x,y) = 0$, and the triangle inequality holds. For the open sets we have only the whole set, how can we get the empty set from the open balls here?
 A: In any pseudometric space $(X,d)$ we define a natural topology
$$\mathcal{T}_d:= \{O \subseteq X\mid \forall x \in O: \exists r>0: B(x,r) \subseteq O\} \text{, where } B(x,r)=\{y \in X: d(x,y) < r\}$$
This can easily be verified to be a topology. $\emptyset$ is always open as we quantify universally over an empty domain and for $O=X$ we can always take $r=1$ for any $x$, e.g. In your example: $B(x,r)=X$ for any $x\in X, r>0$ so $$\forall O \in \mathcal{T}_d: O \neq \emptyset \implies O =X, \text{ so } \mathcal{T}_d = \{\emptyset, X\}$$
i.e. we get the trivial (indiscrete) topology on $X$ for the trivial $0$-pseudometric.
A: The open balls form a base of the topology. The empty set is obtained as the union of the empty family.
A: This pseudometric give rise to the indiscrete or trivial topology. Since $d(x,y)=0$ for all $x,y$, there is one non-empty open ball which is the space $S$ itself. And of course there is the empty ball obtained with a union of an empty family. You can read in the link that in fact every indiscrete space is a pseudometric space with the trivial distance that you have defined, and the converse is also true.
