Let $g_{ij}$ be positive definite metric on a compact manifold $M$. Consider hamiltonian $H=\sum_{ij}\frac{1}{2}g_{ij}(x) p_ip_j$ with $p_i\in\Gamma(T^\star M)$. Then define $E_\lambda=\{(x,p)\in T^\star M\vert H(x,p)=\lambda\}$
$\textbf{Q:}$ $E_\lambda$ is compact. How do I see this obviously? I did the following. Take any open covering of $E_\lambda$ and refine it if necessary s.t. one has trivialization of $T^\star M$. Then project down to $M$ and then lift up to $T^\star M$ and this selects some of the open covering. Since $M$ is compact, $g_{ij}(x)$ has minimal value $g$. Then use $\sum_{ij}g p_ip_j\leq\lambda$ to bound $p_i$ in closed set. Hence, $E_\lambda$ is compact by selecting the finite open covering.
Ref: Riemannian Geometry and Geometric Analysis Jost. Chpt 2 Lie Groups and Vector Bunbldes Pg 70
This is equivalent to showing the sphere bundle $\pi:SM\to M$ is compact for compact manifolds $M$, and your case is just looking at the co-sphere bundle of radius $\sqrt{2\lambda}$, which are equivalent via the bundle isomorphism $F:TM\to T^*M$, $F(v)=2\lambda g(v,\cdot).$
Let $\{x_j,v_j\}\subset SM$ be any sequence. Since $M$ is compact, by possibly passing to a subsequence there exists $x\in M$ such that $x_j\to x$. Let $U\subseteq M$ be a trivializing neighborhood of $x$, i.e., $\pi^{-1}(U)\cong U\times S^{n-1}$. Since $x_j\to x\in U$, for infinitely many $j$, we have that $(x_j,v_j)\in U\times S^{n-1}$. In particular, we have infinitely many $v_j$in the compact set $S^{n-1}$. Again by possibly passing down to a further subsequence there exists some $v\in S^{n-1}$ such that $v_j\to v$. That is, the arbitrary sequence $\{x_j,v_j\}\subset SM$ has an accumulation point in $SM$, thus showing $SM$ is compact.