Let $G$ be Lie group and $K \subset G$ a compact Lie subgroup of $G$. Let $\pi \colon G \to G/K , \quad g \mapsto g.K=[g]$ denote the canonical projection on the quotient and endow $G/K$ with the quotient-topology.
Is $\pi$ then a proper map?
Edit: As I tried to solve it, I used a step I'm not quite sure if it is true.
So let $V \subset G/K$ be a compact subset. Let $(g_j)_j \subset \pi^{-1}(V) =:U$ be sequence. If I now find a convergent subsequence of $(g_j)_j$, then $U$ is compact and $\pi$ proper. Since $\pi(g_j)_j \subset V$ is a sequence and $V$ is compact, there exists a convergent subsequence $(\pi(g_{j_n}))_{j_n}=(g_{j_n}.K)_{j_n} \to g_0.K$. And we assume form now on, that $(\pi(g_j))_j$ is that convergent sequence.
Now comes the part I'm not quite sure: Since $(g_j.K)_j \to g_0.K$, we find a sequence $(k_j)_j \subset K$ such that $g_jk_j \to x_0$ with $x_0.K = g_0.K$ $\textbf{(So do we really find such elements $k_j$?).}$
If we find such a sequence in $K$, we can again use, that $K$ is compact and we find a convergent subsequence $(k_{j_n})_n \to k_0$. Again identify the subsequence with the full sequence and we get: $$g_n = g_n(k_n k^{-1}_n)=(g_nk_n)k_n^{-1} \to x_0k_0^{-1}$$ since the action is continous and so $k_n^{-1} \to k_0^{-1}$.
And we found a convergent subsequence.
Yes, it is proper. Let $C$ be a compact subset of $G/K$ and $(U_i)$ an open cover of $p^{-1}(C)$ without restricting the generality, you can suppose that the adherence $V_i$ of $U_i$ is compact. You can extract a finite family $W_1,...,W_n$ from $p(U_i)$ which covers $C$ where $p:G\rightarrow G/K$ is the natural quotient map. Write $W_i=p(A_i)$ where $A_i$ is an element of the family $(U_i)$. Since the adherence $B_i$ of $A_i$ is compact, the Union $D=\bigcup_i B_iK$ is also compact. You deduce that $p^{-1}(C)$ is a closed subset contained in the compact set $D$ so it is compact.