Bounded subset: Only in metric space or also for premetric space? https://en.wikipedia.org/wiki/Bounded_set defines boundedness of a set as

A subset S of a metric space (M, d) is bounded if there exists r > 0 such that for all s and t in S, we have d(s, t) < r.

after saying that

The word 'bounded' makes no sense in a general topological space without a corresponding metric.

However, I do not see why the triangular inequality is required for $d$. Can I define boundedness the same way if $d$ is a premetric satisfiying

*

*d(x, y) ≥ 0

*d(x, x) = 0

*d(x, y) = d(y, x)

?
 A: No, it does not really make sense for a premetric, at least not with the usual definition of a boundedness notion, in which you want the union of two bounded sets to be bounded.
It does, however, make sense once you do have triangle inequality, e.g. pseudometrics. It also makes sense, for example, once you fix an equivalence class of pseudometrics up to quasi-isometry.
All of this does not invalidate the claim that it does not make sense to talk about bounded sets in a topological space: the point is not that there cannot be a notion of boundedness in a space which has no metric (or even is not metrisable at all), but rather that there is no unique way to assign a notion of boundedness to a pure topological space (i.e. a topological space with no extra structure).
For example, if you look at $\mathbf R$, then you have the standard notion of boundedness that comes from the Euclidean metric, but you also choose a metric which is bounded, making all sets bounded, and the topology won't tell you which ones to choose.
