The image of a finite subgroup of $\text{SU}(2)$ in $\text{SO}(3)$ is a finite subgroup of $\text{SO}(3)$; moreover, the kernel is either trivial or $\{ \pm 1 \}$. But $-1$ is the unique element of order $2$ in $\text{SU}(2)$, so any group of even order contains it.
I claim all the finite subgroups of odd order are cyclic. This follows because the inclusion $G \to \text{SU}(2)$ cannot define an irreducible representation of $G$ (since otherwise $2 | |G|$), hence it must break up into a direct sum of dual $1$-dimensional representations.
So once you know the finite subgroups of $\text{SO}(3)$, you already know the finite subgroups of $\text{SU}(2)$.
Here is my attempt at an answer using the hint from Moishe Kohan.
Let $ H $ be an Ad-irreducible subgroup of a connected group $ G $. If $ H $ is not already an algebraic subgroup then take its Zariski closure $ \overline{H} $. Since $ H $ is Ad-irreducible so is $ \overline{H} $.
An algebraic subgroup of positive dimension and infinite index cannot be Ad-irreducible. Since $ \overline{H} $ is an Ad-irreducible algebraic subgroup then we must have that either $ {\rm dim}(\overline{H})=0 $ or the index $ [G:\overline{H}] $ is finite.
Since $ \overline{H} $ is an algebraic group it can only have finitely many connected components. But since $ \overline{H} $ is discrete every element is a connected component. Thus ${\rm dim}(\overline{H})=0 $ implies $ \overline{H} $ finite which implies $ H $ finite which implies ${\rm dim}(\overline{H})=0 $ (since finite sets are Zariski closed) thus all three statements are equivalent.
Now consider the condition $ [G:\overline{H}] $ is finite. Since we assumed $ G $ is connected that implies $ G/\overline{H} $ is a finite connected manifold thus trivial. So in fact $ G=\overline{H} $. Which is equivalent to saying that $ H $ is Zariski dense.
Thus we have that an ad irreducible subgroup $ H $ of a connected group $ G $ is either finite or Zariski dense.
Stated in a slightly more general way, this shows that an ad irreducible subgroup $ H $ of $ G $ is either finite or $ H $ is Zariski dense in every component of $ G $ that intersects $ H $.
EDIT:
It is interesting to note that for the special case that $ G $ is compact then corollary 3.5 of this paper
https://arxiv.org/pdf/1609.05780.pdf
states that a subgroup $ H $ of a connected compact simple group $ G $ is dense if and only if it is infinite and Ad-irreducible.
So if $ H $ is an ad irreducible subgroup of a connected compact group $ G $ then $ H $ ad irreducible implies $ G $ ad irreducible implies $ G $ simple. So now we have that $ G $ is simple compact and connected. So $ H $ is infinite if and only if it is dense.
In other words, either $ H $ is finite or it is dense and dense implies Zariski dense. Thus corollary 3.5 provides a proof of a stronger but more restricted result (if we strength the assumption to $ G $ compact then in the conclusion of the theorem we can replace Zariski dense with dense(in the manifold topology).
Best Answer
Finite subgroups are always contained in the maximal compact subgroup so if $ G $ has a finite maximal closed subgroup then $ G $ must be compact. Also a maximal subgroup always includes the center so if $ G $ has a finite simple subgroup which maximal then $ G $ must have trivial center in addition to being compact. Thus the only groups which have finite maximal closed subgroups are adjoint groups like $ SO(2n+1), PSO(2n),PU(n) $
Note that a finite simple group $ \Gamma $ is a maximal closed subgroup of $ PU_n $ if and only if the central extension $ n.G $ is a unitary $ 2 $ design as a subgroup of $ SU_n $. See Claim 3 of https://math.stackexchange.com/a/4477296/758507
Some examples of finite simple groups appearing as subgroups of $ PU_n $ are given here
https://mathoverflow.net/questions/414265/alternating-subgroups-of-mathrmsu-n
here
https://mathoverflow.net/questions/414315/finite-simple-groups-and-mathrmsu-n
the references in
https://mathoverflow.net/questions/17072/the-finite-subgroups-of-sun
and here
https://mathoverflow.net/questions/344218/on-the-finite-simple-groups-with-an-irreducible-complex-representation-of-a-give
But when it comes to the maximality of such finite simple subgroups of $ PU_n $ the most useful reference is
https://arxiv.org/abs/1810.02507
which, read correctly, supplies a full classification of maximal closed subgroups of $ PU_d $ that happen to be finite and simple.
The classification consists of a few infinite families of examples of maximal closed subgroups of $ PU_d $ which are finite and simple
$ PU_d $, $ d=\frac{3^k -1}{2} $ and $ d=\frac{3^k +1}{2} $ both have a maximal $ PSp_{2k}(3) $ for $ k \geq 2 $.
$ PU_d $, $ d=\frac{2^k-(-1)^k}{3} $ has a maximal $ PSU_k(2) $ for $ k \geq 4 $
In addition to these, there are a few dimensions $ d $ for which $ PU_d $ has more maximal closed finite simple subgroups than we would expect. These exceptional case are:
$ PU_2 $: $ A_5 $
$ PU_3 $: $ A_6,GL_3(2) $
$ PU_4 $: $ A_7,PSU_4(2) $
$ PU_6 $: $ A_7,PSL_3(4), PSU_4(3) $
$ PU_8 $: $ PSL_3(4) $
$ PU_{10} $: $ M_{11}, M_{12} $
$ PU_{12} $: $ Suz $
$ PU_{14} $: $ ^2 B_2(8) $
$PU_{18} $: $ J_3 $
$PU_{26} $: $ ^3 F_4(2)' $
$PU_{28} $: $ Ru $
$PU_{45} $: $ M_{23},M_{24} $
$PU_{342}$: $ O'N $
$PU_{1333}$: $ J_4 $
All these finite simple maximal closed subgroups of $ PU_n $ lift to finite quasisimple maximal closed subgroups. Some of these quasi simple lifts have simple "section" so to speak and thus correspond to a finite simple maximal closed subgroups of $ SU_n $. Some examples are
$ SU_3 $: $ GL_3(2) $
$ SU_6 $: $ A_7 $