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 $
I claim that $N:=N_{SU(4)}(Sp(2))$ is a maximal subgroup of $Sp(2)$, and that $N = Sp(2) \cup iI Sp(2)$.
To see this, consider the double covering $\pi:SU(4)\rightarrow SU(4)/\{\pm I\}\cong SO(6)$. Note that $-I\in Sp(2)\subseteq SU(4)$, so $\pi|_{Sp(2)}$ is the double covering $Sp(2)\rightarrow SO(5)$. Up to conjugacy, there is an essentially unique $SO(5)\subseteq SO(6)$, the usual block form.
So, instead of studying $Sp(2)\subseteq SU(4)$, we'll study $SO(5)\subseteq SO(6)$ and pull the information back via $\pi$.
Proposition: The only proper subgroup of $SO(6)$ which properly contains $SO(5)$ is $O(5) = \{ \operatorname{diag}(A,\det(A)):A\in O(5)\}$.
Proof: The isotropy action of $SO(5)$ on $S^5 = SO(6)/SO(5)$ is transitive on the unit sphere in $T_{I SO(5)} S^5$, so is, in particular, irreducible. This, then, implies that $SO(5)$ is maximal among connected groups: if $SO(5)\subseteq K\subseteq SO(6)$, then on the Lie algebra level, the adjoint action of $\mathfrak{so}(5)$ would preserve both $\mathfrak{k}$ and $\mathfrak{k}^\bot$, contradicting irreducibility. (Here, I'm using the fact that we can naturally identify the isotropy action of $H$ on $T_{I SO(5)} SO(6)/SO(5)$ with $\mathfrak{so}(5)^\bot\subseteq \mathfrak{so}(6)$.)
Since we now know that $SO(5)$ is maximal among connected subgroups of $SO(6)$, and the identity component of a Lie group is always a normal subgroup, it now follows that $N_{SO(6)}(SO(5))$ is a maximal subgroup of $SO(6)$.
Of course, $O(5)\subseteq N_{SO(6)}(SO(5))$, but why is the reverse inclusion true? Well, every matrix $B\in SO(5)$ fixes the basis vector $e_6$ of $\mathbb{R}^6$, and $\operatorname{span}\{e_6\}$ is the unique subspace of $\mathbb{R}^6$ fixed by all of $SO(5)$. A simple computation reveals that for any $C\in N_{SO(6)}(SO(5))$, that $CSO(5)C^{-1} = SO(5)$ fixes $Ce_6$. It follows that $Ce_6 \in \operatorname{span}\{e_6\}$. Moreover, since $C\in SO(6)$, we must in fact that $Ce_6 = \pm e_6$. In either case, the fact that $CC^t = I$ now implies that $C\in O(5)$. $\square$
Now, let's pull that information back to to better understand $N = N_{SU(4)}(Sp(2))$. It's not too hard to see that $\pi|_N:N\rightarrow N_{SO(6)}(SO(5)$ is a double covering, with $\pi$ mapping $Sp(2)$ and $iI Sp(2)$ to the two different components of $N_{SO(6)}(SO(5))$.
Now, let $g\in N$ be arbitrary. Since $\pi|_{Sp(2)\cup iI Sp(2)}$ is surjective onto $N_{SO(6)}(SO(5))$, there is an $h\in Sp(2)\cup iI Sp(2)$ with $\pi(g) = \pi(h)$. Then $gh^{-1}\in \ker \pi = \pm I$, so $g = \pm I h$. Since both $h, \pm I\in Sp(2)\cup iI Sp(2)$, it follows that $g\in Sp(2)\cup iI Sp(2)$ as well.
Best Answer
WARNING: This answer has lots of true things in it but is also riddled with false claims, I keep meaning to fix it but don't get around to it. $\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}} \DeclareMathOperator\SU{SU}$There exist Lie groups $ \G $ which are not compact or simple but have finite maximal closed subgroups. For example the five element cyclic subgroup $ C_5 $ of $ \mathbb{R}^2 \rtimes C_5 $ described in this answer to Finite maximal closed subgroups of Lie groups.
These sorts of examples are interesting in their own right but are not really what I was thinking about when I asked the question. So from now on I will confine myself to the case that $ \G $ is connected.
In other words I will consider the statement "A connected Lie group $ \G $ has a finite maximal closed subgroup $ G $ if and only if $ \G $ is compact and simple."
The first implication is true.
Claim 1: Let $ G $ be a connected Lie group. Suppose that $ G $ has a finite subgroup which is maximal among the closed subgroups of $ G $. Then $ G $ must be compact and simple.
Proof:
Call such a finite maximal closed subgroup $ \Gamma $. $ \Gamma $ is finite thus compact. So $ \Gamma $ must be contained in some maximal compact subgroup $ K $ of $ G $. Since $ \Gamma $ is maximal then we have that either $ K=\Gamma $ or $ K=G $. Suppose for the sake of contradiction that $ K=\Gamma $. But according to
[https://mathoverflow.net/questions/140622/are-maximal-compact-subgroups-of-connected-groups-connected]
the maximal compact subgroup $ K $ of a connected group $ G $ is always itself connected. So if $ \Gamma=K $ then $ K $ is connected and finite thus trivial. But the trivial subgroup can never be a maximal closed subgroup in a connected Lie group $ G $. The trivial subgroup of a connected Lie group will always be contained in a closed 1-parameter subgroup, which is either a circle or line. So by maximality of $ \Gamma $ that implies the whole group $ G $ is either a circle or a line. But the trivial subgroup is not a maximal closed subgroup of $ S^1 $ or $ \mathbb{R} $. Thus we have a contradiction. So we can conclude that $ K \neq \Gamma $ and thus $$ K=G $$ that is, $ G $ is compact. Now we will show that $ G $ must be simple. Suppose for the sake of contradiction that $ G $ is not simple. Then $ G $ must have some normal subgroup $ N $ such that $$ 0 < dim(N) < dim(G) $$ . So there is a surjective map $$ \pi: G \to G/N $$ Then $ \pi(\Gamma) $ is a maximal closed subgroup of $ G/N $ ( $ \pi(\Gamma) \neq G/N $ since $ dim(\pi(\Gamma))=0< dim(G/N))$ ). Note that $$ \pi^{-1}[\pi(\Gamma)] $$ is a closed subgroup of $ G $ containing $ \Gamma $. But $ \pi^{-1}[\pi(\Gamma)] $ contains $ N $ so $$ 0 <dim(N) < dim(\pi^{-1}[\pi(\Gamma)]) $$ thus $ \Gamma \neq \pi^{-1}[\pi(\Gamma)] $ so by maximality $$ \pi^{-1}[\pi(\Gamma)]=G $$ But that implies that the image of $ \pi $ is contained in $ \pi (\Gamma) $. However that is impossible because $ \pi $ is surjective and $ \pi(\Gamma) $ is zero dimensional while $$ Im(\pi)=G/N $$ is positive dimensional. Thus if a connected Lie group $ G $ has a finite maximal closed subgroup then we can conclude that $ G $ is both compact and simple.
This significantly narrows down the the possibilities for $ G $ to the well known classification of compact simple Lie groups.
However the reverse implication does not hold: $ \operatorname{SU}_{15} $ is an example of a compact connected simple Lie group with no finite maximal closed subgroups.
To see why this is the case it is important to note that
Claim 2: For a compact connected simple Lie group $ \G $, $ G $ is a finite maximal closed subgroup of $ \G $ if and only if $ G $ is Ad-irreducible and $ G $ is a maximal finite subgroup of $ \G $.
The implication Ad-irred+ max finite implies finite and max closed follows from Corollary 3.5 of Sawicki and Karnas - Universality of single qudit gates. The reverse implication, that finite and max closed implies Ad-irred+ max finite follows from the following argument (no it doesn't the reverse implication is false see the EDIT): Let $ \Gamma $ be finite and maximal closed. SSOC that $ \Gamma $ is Ad-reducible. Then $ \Gamma $ acting by conjugation preserves some proper nonzero Lie subalgebra $ \mathfrak{h} $ (EDIT: THIS IS FALSE, $ \mathfrak{h} $ is a sub vector space but not necessarily a subalgebra, so the entire reverse implication falls apart, in particular I believe that the $ SL(2,8) $ subgroup of the exceptional Lie group $ G_2 $ is an example of a finite maximal closed subgroup which is not Ad-irreducible) . So $ \Gamma $ acting by conjugation preserves some connected nontrivial proper subgroup $ H $ of $ G $ with Lie algebra $ \mathfrak{h} $. That implies that $ \Gamma $ normalizes $ H $. That is, $ \Gamma $ is contained in $ N(H) $. But $ \Gamma $ is finite while $ N(H) $ contains the positive dimensional subgroup $ H $. Thus the closed subgroup $ N(H) $ properly contains $ \Gamma $. But $ \Gamma $ is maximal by assumption so that implies $ N(H)=G $. However $ 0 < dim(H) < dim(G) $ and $ H $ is normal, so that contradicts the fact that $ G $ is simple.
Since a finite subgroup of $ \SU_n $ is Ad-irreducible if and only if it is a unitary 2-design we have
Claim 3: $ G $ is a finite maximal closed subgroup of $ \SU_n $ if and only if $ G $ is a maximal unitary 2-group in $ \SU_n $.
By inspecting Theorem 3 of Bannai, Navarro, Rizo, and Pham Huu Tiep - Unitary $t$-groups one immediately determines that $ \SU_{15} $ has no finite maximal closed subgroups.
Some of the main examples of finite maximal closed subgroups of $ \SU_n $ include the normalizer in $ \SU(p^n) $ of an extra-special group $ p^{2n+1} $. Here $ p $ is an odd prime. There is also a similar construction $ p=2 $. Then there are infinite families of examples relating to the Weil module for $ \operatorname{PSp}_{2n}(3) $ and another family related to $ \operatorname{PSU}_n(2) $. Plus many exceptional cases.