About the question of uniqueness of maximal compact subgroups, up to conjugation, I propose a Riemannian geometric approach:

Theorem (É. Cartan). A compact group of isometries of a nonpositively curved complete simply-connected Riemannian manifold has a fixed point.

Proof. (Rough sketch) Consider any orbit. It is compact. By convexity of distance functions (the curvature of the ambient is nonpositive), its center of mass can be defined and it is plainly a fixed
point. QED

Now one uses the fact that the symmetric space of non-compact type $G/K$ is nonpositively curved. If $H$ is a compact subgroup of $G$ then it has a fixed point $gK$ by the theorem, so $g^{-1}Hg$ fixes the basepoint $1K$ and hence is contained in $K$.

**Edit:** on suggestion of Ben, I complete my answer as follows. Let $\mathfrak g = \mathfrak k + \mathfrak p$ be the decomposition of the Lie algebra of $G$ into the eigenspaces of the involution. Since $M$ is complete and nonpositively curved, the Hadamard-Cartan theorem says that the map $\varphi:\mathfrak p \to G/K$ given by
$\varphi(X)=(\exp X)K$ is a smooth covering. Next one sees that $\varphi$ is injective (assume $\varphi(X)=\varphi(Y)$, apply $\mathrm{Ad}$ to both sides and use the uniqueness
of polar decomposition of a matrix, and the semisimplicity of $\mathfrak g$, to deduce that $X=Y$). Hence there are diffeomorphisms $\mathbf R^n\cong\mathfrak p\cong G/K$.
Now it follows rather easily that $\mathfrak p\times K\to G$, $(X,k)\mapsto(\exp X)k$ is a diffeomorhism ($G=\exp[\mathfrak p]\cdot K$ in general for a symmetric space).