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The question shows that an open subset of a manifold is a manifold However I think that statement cannot be extended for Riemannian manifolds because the geodesic curve between two points might not be defined, unless the open subset is geodesically convex.

Is this observation correct?

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    Does a Riemannian manifold require that any two points be connected by a geodesic curve? I thought it was a purely local definition. – 2017-02-10

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Unless your definition of Reimannian manifold is very different from mine, this DOES extend: you define the metric $g'$ on the open subset $U$, i.e., $$ g' : U \to TU^{\otimes 2} $$ to be the restriction of the metric $g$ on $M$ to the subset $U$, i.e., for $p \in U$ and $x, y \in T_p U$ $$ g'(p) (x, y) = g(p) (i_p(x), i_p(y)) $$ where $i_p : T_p U \to T_p M$ is the map induced by the inclusion of $U$ in $M$.

(To clarify: a Riemannian manifold, for me, is a manifold together with a continuously varying metric on the tangent bundle.)

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    Yes, indeed the definition of Riemann manifold is local. But my concern is about the application of the Hopf and Rinow theorem for geodesic completeness of $U$. Suppose that $U$ is connected and every close and bouded subset is compact then there exists a geodesic for each pair of elements in $U$. – 2017-02-10
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    Note that for Hopf-Rinow to apply, you need for $U$ to be complete. But in, say, $\Bbb R^2$, take $U$ to be an open disk. Then $U$ is not complete, as the sequence $(0, 1-\frac{1}{n^2})$ is Cauchy, but its limit, $(0, 1)$ is not in $U$. – 2017-02-10
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    Many thanks for your answer. I will check carefully Hopf-Rinow theorem. – 2017-02-10