I know that $SU(2)$ is homeomorphic and isomorphic to $S^3$. As far as isomorphism is concerned, what I know is that we picked a group structure on $S^3$ that makes it isomorphic to $SU(2)$ by the map $$\phi: \left[ \begin{matrix} x_0 + x_1i & x_2 + x_3 i \\ -x_2 + x_3i & x_0 - x_1i \end{matrix} \right] \to (x_0, x_1, x_2, x_3)$$, but even this I'm not sure of.
But in the case of homomorphism, $S^3$ already inherits it's topology from $\mathbb{R}^4$. Do we define the open sets on $SU(2)$ to be open iff their image is open in $S^3$ under $\phi$?
If that's the case, what's special about $SU(2)$ being isomorphic and homeomorphic to $S^3$? It's like we found a bijective map, and purposely picked a group structure and a topology to make them so.
$SU(2)$ is a subset of $2\times 2$ complex matrices $M_{2\times 2}(\Bbb{C})$, which has its own topology (you can think of flattening it out to $\Bbb{C}^4$). Hence, it inherits a topology from there.
In general, every finite-dimensional real/complex vector space has a unique Hausdorff topology compatible with the vector space operations (and this is also the topology which comes from any choice of norm).
Sometimes, we do define topologies/smooth structures by transport of structure; this isn’t one of those situations.