Is $ SO_3(\mathbb{R}) $ the unique compact connected Lie group acting transitively and faithfully and smoothly on the sphere $ S^2 $?
The point stabilizer of the action will be a closed connected subgroup of codimension 2 (connected because the quotient is simply connected).
@mme says in answer to https://mathoverflow.net/questions/408648/is-so4-a-subgroup-of-su3: "Here is an additional, more straightforward, argument. If H⊂G is codimension 2, then G/H is a surface. An element of g acts trivially on G/H if gg′H=g′H for all g′, which amounts to saying that every conjugate of g lies in H, which amounts to saying that g lies in the intersection of all conjugates of H, which is a normal subgroup of G (call it K). Now G/K is a subgroup of the isometry group of a surface, hence at most 3 dimensional. When g is simple K must be 0-dimensional and dimG≤3."
Maybe that helps?
Here's a sketch of one possible argument:
Let $G$ be a connected compact Lie group acting faithfully, smoothly, and transitively on $S^2$, and let $h$ be an arbitrary Riemannian metric on $S^2$. Let $h'$ be the $G$-invariant "average" of $h$, defined by $$ h'=\int_Gg^*(h)\omega(g) $$ where $\omega$ is the normalized Haar measure on $G$. Since $G$ acts transitively and isometrically, $h'$ has constant curvature, and is thus $(S^2,h')$ is isometric to a round sphere $(S^2,\mathring{h}_r)$. This gives an inclusion $i:G\to SO(3)$, and it's straightforward to argue that $i$ is also an isomorphism.