It's classical (by a back-and-forth argument; see for instance A Shorter Model Theory by Hodges) that any two countable dense unbounded linear orders without endpoints are isomorphic. For example the linear order $\mathbb{Q}$ is isomorphic to $L = \mathbb{Q} \cap (0, 1)$, and they are homeomorphic under their order topologies.

So it will suffice to exhibit a homeomorphism $f: L \times L \to L$. One method might be to interleave two eventually periodic decimal expansions (forbidding infinite tails of 0's, say), using one expansion at the odd places and the other at the even places. The result is again eventually periodic, and is bicontinuous.

Edit: The following is a second attempt to repair problems in earlier proposed solutions, as pointed out by Gerald Edgar in comments. Hopefully this time I've gotten it right this time.

It's classical (by a back-and-forth argument; see for instance A Shorter Model Theory by Hodges) that any two countable dense unbounded linear orders without endpoints are isomorphic. For example the linear order $\mathbb{Q}$ is isomorphic to $L = \mathbb{Z}_{(2), (5)} \cap (0, 1)$, referring here to the integers localized at the primes 2 and 5 (i.e., rationals whose decimal expansions don't have an infinite tail of 9's or 0's -- this is for technical reasons). $L$ and $\mathbb{Q}$ are homeomorphic under their order topologies.

Let $f: L \times L \to \mathbb{Q} \cap (0, 1)$ be the map that takes a pair of elements $\alpha, \beta \in L$ and forms a rational number by interleaving their decimal expansions, with the decimal expansion of $\alpha$ appearing in odd places and that of $\beta$ in the even places. Let $I$ be the image of $f$. The map $f: L \times L \to I$ is a homeomorphism, and $I$ is again a countable dense linear order without endpoints, hence homeomorphic to $\mathbb{Q}$ again. Then the evident composite

$$\mathbb{Q} \times \mathbb{Q} \cong L \times L \stackrel{f}{\to} I \cong \mathbb{Q}$$

is a homeomorphism $\mathbb{Q} \times \mathbb{Q} \cong \mathbb{Q}$.

It's classical (by a back-and-forth argument; see for instance A Shorter Model Theory by Hodges) that any two countable dense unbounded linear orders without endpoints are isomorphic, so that for example the linear order $\mathbb{Q}$ is isomorphic to $\mathbb{Q}(\sqrt{2})$ ordered as usual. So they are homeomorphic under their order topologies. Then $\mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}(\sqrt{2})$, sending $(r, s)$ to $r + s\sqrt{2}$ is a homeomorphism. Compose these two homeomorphisms to establish a homeomorphism $\mathbb{Q} \times \mathbb{Q} \cong \mathbb{Q}$.

There are other methods, based on for example putting pairs of finite continued fractions together to form a continued fraction of greater size, that could be used (after some slight finagling).

It's classical (by a back-and-forth argument; see for instance A Shorter Model Theory by Hodges) that any two countable dense unbounded linear orders without endpoints are isomorphic. For example the linear order $\mathbb{Q}$ is isomorphic to $L = \mathbb{Q} \cap (0, 1)$, and they are homeomorphic under their order topologies.

So it will suffice to exhibit a homeomorphism $f: L \times L \to L$. One method might be to interleave two eventually periodic decimal expansions (forbidding infinite tails of 0's, say), using one expansion at the odd places and the other at the even places. The result is again eventually periodic, and is bicontinuous.