As reported here I know that:
Let $(M,g)$ be a complete connected riemannian manifold, let $S \subset M$ be connected and closed subset of M, if it is also an embedded submanifold of M, then $(S, i^*g)$ is complete.
My question is if it is still true when $S$ is and immersed (but not embedded) submanifold of M. More explicit:
Let $(M,g)$ be a complete connected riemannian manifold, let $S \subset M$ be connected and closed subset of M, if it is also an immersed submanifold of M, then $(S, i^*g)$ is complete (as a connected riemannian manifold, i.e. in the Hopf-Rinow sense)?
I think that the answer is no. In particular, since the inclusion is smooth for all smooth curve $\gamma:I \to S$, can I view it as $\beta=i \circ \gamma $ smooth curve on $M$ and $L_g(\beta) = L_{i^*g}(\gamma)$ then $d_g|_S \leq d_{i^*g}$. Thus every S-cauchy sequence is M-cauchy and the M-completeness with the S-closure find a candidate limit point in S. However because S can have a topology different from the subspace topology, I think that this candidate in general will not be an S-limit point for the sequence.
I try to find an explict counterexample, but I don't know an example of immersed closed submanifold that is not compact.
