This may be a very silly question but let us consider a connected Riemanian manifold $(M,g)$ and a subset $O\subset M$. Can we have $O$ geodesic complete (in the sense of all geodesics linking two points in $O$ can be extended and still remains in $O$) without $O=M$ ?
Can we have $dim(O)=dim(M)$ with $O\subsetneq M$ ?
As noted in the comments, totally geodesic submanifolds give examples with $\dim O < \dim M$. If you want $O$ to have full dimension then the only examples are (unions of) the connected components of M:
Let $M$ be a complete connected manifold and $O$ be a complete open subset. If $x\in O$ then for any point $p \in M$, let $\gamma:[0,1]\to M$ be the minimising geodesic joining $x$ to $p$ (guaranteed by Hopf-Rinow). Since $O$ is open, there is some $\epsilon>0$ such that the restriction of $\gamma$ to $(-\epsilon,\epsilon)$ is a geodesic in $O$. Therefore the full extension $\gamma$ (and in particular $p$) must lie in $O$ by the completeness of $O$.