It is possible to have nonstandard models of PA where the natural numbers are different.
The definition of a topology requires a notion of finiteness.
What happens if we use a nonstandard model of arithmetic to say what finiteness means in the definition of topology?
On
The definitions of topology require some notion of set theory. Therefore we need to work in a theory which is stronger than that of $\sf PA$.
Mostly we work in $\sf ZFC$, or some variation of it. The important part is that these theories prove the consistency of $\sf PA_2$, that is the second-order theory of $\sf PA$, which is categorical. So internally to a universe of $\sf ZFC$ there is no such thing as a non-standard integer.
Of course, there are many different ways of defining finiteness (especially in $\sf ZFC$), which do not make any appeal to the integers themselves. But the point is that topology is working within the universe of a theory that proves that all the integers are standard (again, internally, it might be the case that the model itself does have non-standard integers, but it cannot know about it from an internal point of view).
It turns out that we needn't use arithmetic to define finiteness at all. For example, a set $S$ of sets is finite (in the sense we're used to) if and only if each non-empty subset of the power set of $S$ has a minimal element with respect to inclusion ($\subseteq$). A general set $X$ (not necessarily a set of sets) is finite if and only if every injection $X\to X$ is a surjection. Alternately, $X$ is finite if and only if every inductive family of subsets of $X$ contains $X$. An inductive family of subsets of $X$ is a set $\mathcal{A}$ of subsets of $X$, such that $\emptyset\in \mathcal{A}$, and such that for all $A\in\mathcal{A}$ and all $x\in X$ we have $A\cup\{x\}\in\mathcal{A}$.