Consider a topological space $(X_1, \tau_1)$ and a set $X_2$. Suppose as well that there existed some bijection $h$ from $X_1$ to $X_2$ (and viceversa).
Now suppose that we attempted to construct an "induced" topology for $X_2$, $\tau_2$, by considering a subset of $X_2$ to be an open set in $\tau_2$ if and only if it is the image of an open set in $\tau_1$ under the action of the bijection $h$. Would this technique always generate a topology for $X_2$? If not, are there any simple restrictions on the topological characteristics of $(X_1,\tau_1)$ that would make this work (i.e., compactness)?
Yes. Your definition means that $$\tau_2 = \{h(U) \mid U \in \tau_1\} .$$ It is very easy to verify that $\tau_2$ is a topology on $X_2$. This topology makes $h$ a homeomorphism ($h$ is open by definition and continuous because $h^{-1}(h(U)) =U$) and it is the only topology on $X_2$ with this property. Indeed, let $\tau$ is a topology on $X_2$. Then $h : (X_1,\tau_1) \to (X_2,\tau)$ is open iff $\tau_2 \subset \tau$ and continuous iff $\tau \subset \tau_2$.