I'm looking for a proof (that I can understand) of the following fact: If $K$ and $G$ are Lie groups, and $K$ is compact, then nearby homomorphisms $K\to G$ are conjugate.

That is, if $\mathrm{Hom}(K,G)$ is the set of Lie group homomorphisms, endowed with a suitable topology (I'd like to say compact-open), then the orbits of the conjugation-by-$G$ action on it are open. (Note that there are obvious conterexamples if $K$ is not compact.)

This is referred to in The space of Lie group homomorphisms. A reference is given there to Connor-Floyd, Differentiable Periodic Maps, Ch. VIII, Lemma 38.1.

Following the reference, we see that Connor and Floyd derive this as an easy consequence of a theorem from Montgomery-Zippin, Topological Transformation Groups, p. 216. That is, by thinking about the graph $K\to K\times G$, of a homomorphism, the statement can be deduced from the following:

• If $K\subseteq G$ is a compact subgroup, then there exists a neighborhood $U$ of $K$ in $G$ such that for any subgroup $H\subset U$, there exists $g\in G$ such that $gHg^{-1}\subseteq K$. (I.e., all subgroups "close" to a compact subgroup are conjugate to a subgroup of it.)

Montgomery-Zippin's proof is an exercise involving geodesics in symmetric spaces, which is opaque to me and will probably always remain so. (They have statements such as: "there exists a neighborhood $U$ of $x$ such that for any geodesic in $U$, for any points $a,b,c$ in that order along the geodesic, $d(x,b)< \mathrm{max}(d(x,a),d(x,c))$" (quoting from memory, don't take it literally). I'm just a simple algebraic topologist, and sort of thing goes right over my head.)

Can anyone describe a more modern proof, or give a reference? I'm imagining such a proof will be an exercise involving the exponential map. (In fact, it seems easy to prove that any subgroup "close" to the identity is trivial in just this way.)