The Möbius strip (without boundary) $ S $ can be realized as a regular surface of $ R^3 $ (regular surface is meant in the sense of do Carmo's book). Using a 'manifold' language it can be proved that $ S $ is an embedded submanifold of $ R^3 $. Therefore it is for itself a smooth manifold (that we realize as embedded into $ R^3 $).
My trouble is that we cannot obtain the Möbius strip with boundary as an embedded subanifold of $ R^3 $. If we want to realize the Möbius band with boundary it seems natural to consider the topological closure of $ S $ as a subset of $ R^3 $ and then proving that this closed subset is an embedded submanifold with boundary into $ R^3 $. However this argument should fail since we know that no closed subset of $ R^3 $ cannot be homeomorphic to a non orientable $ 2 $-manifold.
My question is : why does this argument should fail? I haven't checked the detail but i'm looking for an intuitive answer for it.
N.B: an embedded submanifold $ M \subset R^3 $ is defined as a subset such that for every $ p \in M$ there exists a diffeomorphism $ \phi:U \rightarrow V $, where $ p \in U $, $ \phi(p) =0 $ and $ U,V $ are open subsets of $ R^3 $ such that $ \phi(U \cap M)=V \cap R^2 $