Let $G$ be a Lie group acting smoothly and properly on a smooth manifold $M$. Denote the action by $$\psi:G\times M\longrightarrow M,\quad \psi(g,p)=\psi_g(p).$$ Then, there is a natural action of $G$ on the tangent bundle of $M$ given by $$G\times TM\longrightarrow TM,\quad (g,v)\longmapsto d\psi_g(v).$$
Question: Is the action proper?
Answer: Yes.
Proof: One of the equivalent definitions of proper action (see Lee, Introduction to Smooth Manifolds, p.543) is:
$\psi$ is proper if and only if for any sequences $p_i\in M$ and $g_i\in G$ such that both $p_i$ and $g_i\cdot p_i$ converge, a subsequence of $g_i$ converges.
Now, let $(p_i,v_i)\in TM$ and $g_i\in G$ be such that $(p_i,v_i)$ and $d\psi_{g_i}(p_i,v_i)$ converge. Then, the projection of the sequence $d\psi_{g_i}(p_i,v_i)$ to $M$ is the sequence $g_i\cdot p_i$, which converges by continuity. Similarly, $p_i\in M$ converges. Hence a subsequence of $g_i$ converges.
