A closed subset of $U$ open in $M$, a manifold, is closed in M?

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Let $M$ be a $C^{\infty}$ manifold. Let $U$ be an open subset of $M$. Now take a closed subset (with respect to the subspace topology on $U$) $V \subseteq U$. Does is then follow that $V$ is closed in $M$? Thank you.

PS I would further like to add more conditions as it appears that I may have simplified the problem too much as shown by Kavi Rama Murthy below. We further let $(U, \phi)$ be a local chart with $p \in U$, and $V = \phi^{-1}(\overline{B(\phi(p), \varepsilon)})$ for $\epsilon > 0$ small so that $\overline{B(\phi(p), \varepsilon)} \subseteq \phi(U)$.

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Under the additional assumptions it is true. $V$ is homeomorphic to a closed ball in euclidean space and hence compact. Since $M$ is Hausdorff this implies that $V$ is closed in $M$.

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No. Take $M=\mathbb R, U=V=(0,1)$.