$ SO_2(\mathbb{R}) $ does not have any maximal closed subgroups.
The maximal closed subgroups of $ SO_3(\mathbb{R}) $ are $ O_2(\mathbb{R}) $ as well as the two finite groups $ S_4 $ and $ A_5 $.
$ SO_4(\mathbb{R}) $ double covers $ SO_3(\mathbb{R}) \times SO_3(\mathbb{R}) $. So the six maximal closed subgroups of $ SO_4(\mathbb{R}) $ are the lifts through this double cover of the three subgroups of the form $ H \times SO_3(\mathbb{R}) $ and the three subgroups of the form $ SO_3(\mathbb{R}) \times H $, where $ H $ is one of the three maximal closed subgroups of $ SO_3(\mathbb{R}) $.
What are the maximal closed subgroups of $ SO_5(\mathbb{R}) $?
https://arxiv.org/pdf/math/0605784.pdf
classifies all maximal closed subgroups of $ SO_5 $ whose identity component is not simple (here trivial counts as simple). According to this paper, pages 1024-1025, the maximal closed subgroups of $ SO_5 $ of this type are
$$
S(O_3 \times O_2 )
$$
with two connected components and
$$
S(O_4 \times O_1) \cong O_4
$$
also with two connected components.
Next, we consider maximal closed subgroups with nontrivial simple connected component.
By dimension, such a subgroup would be isogeneous to $ SU_2$ or $ SU_3 $.
There are no 5d irreps of $ SU_3 $ since the dimension of $ SU_3 $ irreps are given by the formula
$$
\frac{(m_1+1)(m_2+1)(m_1+m_2+2)}{2}
$$
I think that the irreducible real 5 dimensional irrep of
$$
SO_3(\mathbb{R})
$$
is the identity component of a maximal closed subgroup of $ SO_5 $. Perhaps it is the entire maximal group? Or perhaps the normalizer $ N(SO_3(\mathbb{R})) $ is slightly larger?
Finally, what (if any) are the finite maximal closed subgroups of $ SO_5(\mathbb{R}) $?
$$
S_6
$$
of order $ 720 $ is a finite maximal closed subgroup. There is a faithful 4d complex irrep of $ 2.S_6 \cong Aut(SL_2(9)) $ which is quaternionic (Schur indicator -1) so $ 2.S_6 $ is a subgroup of $ Sp_2 $. Thus $ S_6 $ is a subgroup of $ Sp_2/\pm I \cong SO_5(\mathbb{R}) $. One can also verify directly that $ S_6 $ has a faithful 5d real irrep. Since this $ S_6 $ subgroup is irreducible and not a subgroup of $ SO_3 $ it it not contained in any infinite closed group. Since $ S_6 $ is maximal among the finite subgroups that is enough to conclude that $ S_6 $ is a finite maximal closed subgroup of $ SO_5 $.
The other finite maximal closed subgroup of $ SO_5 $ is the group of signed permutation matrices of determinant 1
$$
W(D_5)
$$
of order $ 5!2^5/2=1920 $ also known as the Weyl group of $ D_5 $. This is related, in a slightly convoluted way, to the PerfectGroup(1920,6) subgroup of $ Sp_2 $.
This group is irreducible and not a subgroup of $ SO_3 $. Thus all that remains to show is that it is maximal among the finite subgroups of $ Sp_2 $.
$ 2.W(D_5) $ is the intersection of $ Sp_2 $ with group XXVII from https://arxiv.org/abs/hep-th/9905212. XXVII isn't quite maximal among the finite subgroups of $ SU_4 $, since it is contained in group XXX. However $ 2.W(D_5) $ is maximal finite in $ Sp_2 $ because group XXX cannot be contained in $ Sp_2 $ by part (i) of theorem 8 of https://core.ac.uk/download/pdf/82740228.pdf
(where $ 2.W(D_5) $ is referred to as calligraphic B*)
To summarize, the full list of maximal closed subgroups of $ SO_5 $ is
\begin{align*}
& S(O_3 \times O_2) \\
& S(O_4 \times O_1)\cong O_4 \\
& SO_3 \\
& S_6 \\
& W(D_5)
\end{align*}
I'm interested in any references or thoughts on the correctness of this list. I'm especially curious about the maximality of the 5d irrep of $ SO_3 $, and thoughts on the finite maximal subgroups. Note that $ SO_3 \times SO_2 $ in $ SO_5 $ lifts to $ U_2 $ in $ Sp_2 $ and $ O_4 $ in $ SO_5 $ lifts to $ O_4 $ in $ Sp_2 $.
Note on $ n=6 $ case: Since $ SO_6(\mathbb{R})\cong SU_4/\pm I $ then the maximal closed subgroups of $ SO_6 $ are exactly the quotient by $ \pm I $ of the groups given in Maximal Closed Subgroups of $ SU_4 $
Namely, the maximal closed subgroups of $ SU_4 $ and their corresponding subgroups in $ SO_6 $ are
\begin{align*}
& N(T) \twoheadrightarrow N(T) \cong (SO_2 \times SO_2 \times SO_2) \rtimes S_3 \\
& S(U_3 \times U_1) \cong U_3 \twoheadrightarrow U_3 \\
& N(S(U_2 \times U_2)) \twoheadrightarrow S(O_2 \times O_4) \\
& N(SU_2 \otimes SU_2) \twoheadrightarrow N((SO_3 \times SO_3)) \\
& N(Sp_2) \twoheadrightarrow S(O_5 \times O_1)\cong O_5 \\
&4 \circ_2 2.A_7 \twoheadrightarrow 2 \times A_7 \\
&4 \circ_2 Sp_4(\mathbb{F}_3) \twoheadrightarrow 2 \times PSp_4(\mathbb{F}_3) \\
&N(2^{2(2)+1}) \twoheadrightarrow W(D_6)
\end{align*}
where in the second line $ U_3 $ is the image of the standard embedding of $ U_n $ into $ SO_{2n}(\mathbb{R}) $, the third group has cyclic 2 component group generated by minus the identity, and for the fourth group the normalizer has cyclic 4 component group generated by
$$
\begin{bmatrix} 0 & I\\ -I & 0 \end{bmatrix}
$$
where here $ I $ is the $ 3 \times 3 $ identity matrix.
Note that throughout we have used freely the commuting diagram where
$$
Sp_2 \hookrightarrow SU_4
$$
double covers
$$
SO_5 \hookrightarrow SO_6
$$
Best Answer
The final answer is that $SO(3)$ (embedded into $SO(5)$ as an irrep) is maximal.
I'll first prove that $SO(3)$ is maximal among connected Lie groups, and then I'll worry about components after that.
So, let's assume $SO(3)\subseteq K\subseteq SO(5)$ with $K$ connected. We can assume $SO(3)\neq K\neq SO(5)$. Then counting dimensions and rank shows that $K$ is, up to cover, one of $SO(3)\times S^1, SO(3)\times SO(3)$, or $SU(3)$. You have already shown $SU(3)$ is not contained in $SO(4)$, so it must be one of $SO(3)\times S^1$ or $SO(3)\times SO(3)$. In any case, $K$ has a cover of the form $K' = SU(2)\times S^1$ or $SU(2)\times SU(2)$.
The irreps of $K'$ are tensor products of irreps of the factors. The factor $SU(2)$ has a unique irrep in each (complex) dimension, which alternates between real and quaternionic. The irreps of $S^1$ are all one dimensional, and classified by integers. They are complex, except the trivial rep is real.
Let's start with $K' = SU(2)\times S^1$. Since the rep a) be $5$-dimensional and b) contain $(\mathbf{5}\otimes \rho)$ as a sub rep (where $\mathbf{5}$ denotes the $5$-dim irrep of $SU(2)$ and $\rho$ denotes an arbitrary irrep of $S^1$), it follows that the rep of $K'$ given by $K'\rightarrow K\subseteq SO(5)$ is of the form $\mathbf{5}\otimes \rho$. If $\rho$ is trivial, then the covering $K'\rightarrow K$ has kernel containing $\{e\}\times S^1$, so is not finite, a contradiction. If $\rho$ is non-trivial, the rep is a tensor product of an orthogonal rep with a complex rep, so is complex (not orthgonal), again, giving a contradiction. Thus, this case cannot occur.
Let's next assume $K' = SU(2)\times SU(2)$. Unlike the previous case, we cannot simply assert that such a rep must contain $\mathbf{5}\otimes \rho$ because the $SO(3)$ we care about a priori could be diagonally embedded, or something like that.
So, let's just find all almost-faithful $5$-dim real reps of $K'$. If such a rep includes a sub rep of the form $\mathbf{1}\otimes \mathbf{n}$, of dimension $n\geq 2$, it must also include one of the form $\mathbf{m}\otimes \mathbf{n'}$ with $m\neq 1$ and $mn' = 5-n$. From here, it's easy to see that $n' = 1$. From here, it follows that precisely one of $n,m$ is odd, which then implies the rep is complex. (A real reap is a sum of real reps together with a sum of other irreps paired with their duals.)
So, the only reps we need to consider have the form $\mathbf{m}\otimes \mathbf{n}$ with both $m,n\geq 2$. Then it's clear the only option is $(\mathbf{2}\otimes \mathbf{2}) \oplus (\mathbf{1}\otimes\mathbf{1}).$ This rep (which corresponds to the usual $SU(2)\times SU(2)\rightarrow SO(4)\subseteq SO(5)$ is obviously reducible. This contradicts the fact that the $SO(3)\subseteq K$ action is irreducible. This concludes the case where $K$ is connected.
So, $SO(3)$ is maximal among connected groups, but what about components? Well, as in your previous questions, it follows now that $N:=N_{SO(5)}(SO(3))$ is a maximal subgroup of $SO(5)$ containing $SO(3)$. Let's see why $N = SO(3)$.
So, suppose $g\in N$. Then conjugation by $g$ induces an isomorphism of $SO(3)$. All isomorphisms of $SO(3)$ are inner, so there is an $h\in SO(3)$ with $hAh^{-1} = gAg^{-1}$ for all $A\in SO(3)$. Said another way, $h^{-1}g$ centralizes $SO(3)$.
Now, think of $SO(3)$ as acting on $\mathbb{C}^5$ as a complex rep, Schur's lemma tells us the the isomorphisms of the rep are $\mathbb{C}$-multiples of the identity matrix $I$. But anything which centralizes $SO(3)$ is an isomorphism of the rep. Thus, $h^{-1}g = \lambda I$ for some $\lambda \in \mathbb{C}$. As the only multiple of the identity in $SO(5)$ is the identity, we conclude $h^{-1}g$ is the identity, That is, $h = g$. This proves $N\subseteq SO(3)$, so $SO(3)$ is, in fact, a maximal subgroup of $SO(5)$.