What is the topology of the hyperreal line?

I believe there isn't a single thing we can point to as "the" right topology on the hyperreals, but there are a few natural candidates:

• The interval topology was discussed in the comments.

• We can take as basic opens the intervals $(r-\epsilon, r+\epsilon)$ with $r\in {}^*\mathbb{R}$ and $\epsilon\in\mathbb{R}_{>0}.$ Unions of such intervals are called real open sets; unfortunately the real open sets are not closed under intersection, and the induced topology is not well-behaved.

• We can replace $(r-\epsilon, r+\epsilon)$ with $((r-\epsilon, r+\epsilon))=\{x: x$ is well inside $(r-\epsilon, r+\epsilon)\}.$ (This just means that we demand $x$ not be infinitesimally close to $r-\epsilon$ or $r+\epsilon$, in addition to $x\in(r-\epsilon, r+\epsilon)$.) These are the $S$-neighborhoods, and they induce the $S$-topology. This seems to be the nicest standard topology on the hyperreals.

See Robert Goldblatt's book, esp. chapters 10 and 11.

The "sameness" of the standard and nonstandard models is about elementary equivalence, not homeomorphism; by the transfer principle, $\mathbb{R}$ and ${}^\star \mathbb{R}$ have exactly the same internal properties.

That is, the most well-behaved topology on ${}^\star \mathbb{R}$ is the internal topology given by the transfer of the usual topology on $\mathbb{R}$, but it's only well-behaved internally. For example, it only has the internal least upper bound property: every bounded, nonempty, internal subset has a least upper bound. External subsets can go either way.

Whether or not that internal topology is an external topology is a nontrivial question I'm not prepared to answer; even at the most naive level of taking some formulation of topology verbatim, the answer depends on which formulation of topology you use!