Disk Vector Bundle is a Subbundle I'm having difficultly proving that the disk bundle $D(E)$ of a vector bundle $p: E \to B$ is a subbundle. 
The definition I have for $D(E)$ is $D(E) = \{e \in E \ |\ \|e\| \leq 1 \}$
The criteria for $D(E)$ to be a subbundle are:
(i) $D(E)$ is a subspace of $E$
(ii) $D(E) \cap p^{-1}(b)$  is a vector space for ever $b \in B$
(iii) $p|_{D(E)} : D(E) \to B$ is a vector bundle. 
My Proof thus far:
(i) is clear from the definition of $D(E)$.
(ii) We know by the definition of vector bundle that $p^{-1}(b)$ is a vector space. However there is no such constraints on $E$ so I cant say that $D(E)$ is a vector space. And this is where I'm having my first issue. 
(iii) Not a single clue of how to show this. 
 A: If $E\overset{\pi}\to B$ is a bundle over base $B$, and $F$ is a subspace of $E$, then $F$ is a subbundle if $F$ is also a bundle over $B$ (so we have $F\overset{p}\to B$) in a compatible way with the bundle structure. In other words so that each fiber of $F$ $p^{-1}(b)$ includes into the corresponding fiber $\pi^{-1}(b)$ of $E$. More abstractly, the following diagram must commute.

Being a vector bundle is just to say that each fiber is a vector space, in a bundle compatible way. While in some contexts, if someone were to speak of a subbundle of a vector bundle, it may be reasonable to assume they mean vector subbundle, here that is not intended. This subbundle is only a fiber subbundle, not a vector subbundle, since a disk is not a vector space.
So what do you have to prove? Only that for each fiber $p^{-1}(b),$ the fiber of $D(E)$ over $b$ which is $\{e\in E\mid \lVert e\rVert \leq 1 \text{ and } p(e)=b\}$ includes into $p^{-1}(b)$. Which is true by construction.
You also have to check the local triviality condition, which I think is your (iii). In other words, there is a neighborhood $U$ of each point in the base where $D(E)|_U = U\times D(V)$. This is easily seen to be furnished by $D(E_U).$ In other words, the disk bundle of a locally trivial patch of the vector bundle gives a locally trivial patch of the disk bundle.
But since $D(E)$ is not expected to be a vector bundle, we do not need to check the vector space axioms.
