In one of the books on algebraic topology (I don't remember exactly which one) there was an exercise to build an example of two topological spaces having two continuous bijections between them which are not homotopy equivalent. To be honest, this exercise confuses me a little because, as I understand, each pair of homeomorphic spaces is homotopy equivalent by construction. On the other hand, the existence of bijective continuous mapping between spaces automatically provides their homeomorphism (correct me if I'm wrong). Thus, in this logic, if we have two continuous bijections between topological spaces this will inevitably lead to the homotopy equivalence between them. I guess, however, that there is a simple counterexample related to the discrete topologies which breaks such a reasoning (see, for example, this: Is a bijective homotopy equivalence with bijective homotopy inverse a homeomorphism?), but I have certain difficulties in discovering it. Is there any suggestion?
The interval $[0,1)$ admits a continuous bijection to the circle (use $\exp(i 2\pi t)$ as your map). But the two spaces are not homotopy equivalent since the circle is not simply connected.
Edit: If you want to have more than one continuous bijection, just compose my map with a rotation of the circle.