Compact subgroups of SL_n are conjugate into SO(n)

Idea. A compact group preserves some inner product. In an orthonormal basis for that inner product, $K$ acts by orthogonal matrices.

Step 1: an invariant inner product. $K$ is a compact topological group, so it carries a left-invariant probability measure $\mu$ (Haar measure). For $x,y\in\mathbb{R}^n$ define
$$\langle x,y\rangle'=\int_K\langle gx,gy\rangle\,d\mu(g),$$
where $\langle\cdot,\cdot\rangle$ is the standard inner product. The integrand is continuous in $g$ (compact domain, so integrable). $\langle\cdot,\cdot\rangle'$ is bilinear, symmetric, and positive definite: $\langle x,x\rangle'=\int\|gx\|^2d\mu>0$ for $x\ne0$ since $\|gx\|>0$.

Step 2: it is $K$-invariant. For $h\in K$,
$$\langle hx,hy\rangle'=\int_K\langle ghx,ghy\rangle\,d\mu(g)=\int_K\langle g'x,g'y\rangle\,d\mu(g')=\langle x,y\rangle'$$
by right invariance of the Haar measure on a compact group (compact groups are unimodular, so left and right invariant measures coincide).

Step 3: change of basis. There is a symmetric positive definite matrix $S$ with $\langle x,y\rangle'=\langle Sx,y\rangle$. Let $P=S^{1/2}\in GL_n(\mathbb{R})$, so $\langle x,y\rangle'=\langle Px,Py\rangle$. For $h\in K$ the invariance reads $\langle Phx,Phy\rangle=\langle Px,Py\rangle$ for all $x,y$. Putting $x'=Px$, $y'=Py$ gives
$$\langle PhP^{-1}x',PhP^{-1}y'\rangle=\langle x',y'\rangle\quad\text{for all }x',y',$$
so $PhP^{-1}\in O(n)$.

Step 4: determinant. $\det(PhP^{-1})=\det h=1$ since $K\subset SL_n$. So $PKP^{-1}\subset SO(n)$.

Remark. The same proof shows that any compact subgroup of $GL_n(\mathbb{R})$ is conjugate into $O(n)$. It gives a "maximal compact subgroup" theorem: $SO(n)$ is a maximal compact subgroup of $SL_n(\mathbb{R})$, unique up to conjugation (Cartan-Iwasawa-Malcev).