Suppose $\{x_n\}_{n=1}^{\infty}$ is a bounded sequence in a finite dimensional Riemannian manifold. Can I say that this set is relatively compact?
In $R^n$, this is true by Bolzano-Weierstrass theorem. I cannot think of a way to transfer that argument to any finite dimensional Riemannian manifold.
Let's assume that $M$ is connected. You cannot transfer that argument as this is not true. For example if $(M, g) = \{x\in \mathbb R^n: |x| <1\}$ with the Euclidean metric, one can easily construct a bounded sequence which does not converge.
One need to assume that $M$ is complete (with respect to the metric $d$ induced from the $g$). By Hopf-Rinow theorem, this is the same as that the exponential map $\exp_p$ is defined on the whole $T_pM$ for all $p\in M$. Now let $\{x_n\}$ be a bounded sequence in $M$. Then there is $p\in M$ so that $d(x_n, p) <C$ for all $n$. Thus we can find $v_n\in T_pM$ so that $\exp_p(v_n) = x_n$ and $\| v_n\| < C$. By the usual Bolzano-Weiestrauss theorem on $\mathbb R^n$, $v_n \to v$ for some $v\in T_pM$. By continuity of $\exp_p$, $\{x_n\}$ converges to $\exp_p(v) \in M$.