Why must subbundle be an embedded submanifold? In John Lee's smooth manifold, pg 264, or 199 in this version, he defines 

Let $\pi:E \rightarrow M$ be a vector smooth bundle. A subbundle $\pi|_D:D \rightarrow M$ is called a smooth subbundle if it is smooth vector bundle and an embedded submanifold with or without boundary. 


Being embedded in Lee's definition has two conditions: 


*

*It is a topological embedding. 

*It is a smooth immersion. That is $di_p :T_pD \rightarrow T_pE$ is injective for all $p \in D$. 



$D$ can be given  subspace topology. 1. is then satisfied. 
But why do we require in the definition that it is a smooth immersion?  Is this condition additonal? Or could actually be deduced? 
 A: Requiring that a subbundle be embedded determines its topological and smooth structure uniquely. In fact it is generally true that if $S\subseteq M$ then there is at most one topological and smooth structure with respect to which the inclusion is a smooth embedding. See Theorem 5.31 on page 114 of Lee's Smooth Manifolds.
The requirement is also natural for the reason pointed out by Matt. It allows us to think of the tangent space $T_pD$ of a subbundle as a subspace of the tangent space $T_pE$ of the bundle.
At the same time that embedding immediately gives us uniqueness of smooth structure on a subbundle (although Theorem 5.31 is actually fairly nontrivial), I also believe that uniqueness follows even without the embedding condition. Suppose $D\subseteq M$ has two different smooth structures under which it is a subbundle. These two smooth structures give two different local frames for $D$ in a neighborhood of any point. Composing with the differential of the smooth inclusion map, these smooth frames are also smooth in $M$. Then vector fields in either frame can be expressed as a smooth linear combination of vector fields in the other frame, meaning that the two smooth structures have to be the same.
I'm also fairly sure that any smooth (not necessarily embedded) subbundle must be embedded anyway. This would mean that the embedded condition is redundant, but it doesn't hurt and does make certain things clear from the get-go.
