Cohomology of the Thom Space of a Vector Bundle

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The Thom space $T(E)$ of a vector bundle $E \to B$ with metric is defined as $D(E)/S(E)$, where $D(E)$ denotes the disk bundle and $S(E)$ denotes the sphere bundle of $E \to B$.

I've been trying to prove that $H^{*}(T(E))$ is isomorphic to $H^{*}(E,E_{0})$, where $E_{0}$ is the complement in $E$ of the zero section. I already proved that $H^{*}(E,E_{0})$ is isomorphic to $H^{*}(D(E),S(E))$ (using the long exact sequence of a pair and that $D(E)$ is a deformation retract of $E$ such that if $r$ is such deformation retraction then $r(E_0) \subset S(E)$.)

I think I have to use excision to the pair $(D(E), S(E))$ in order to realize the cohomology $H^{*}(D(E),S(E))$ as the cohomology $H^{*}(T(E))$ but I can't realize how to do it. Any suggestion?

Thanks in advance!

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You did the hardest part already. By dualizing Proposition 2.22 in Hatcher or referring to Relative Cohomology Isomorphic to Cohomology of Quotient we get that $$ H^*(D(E),S(E)) = \tilde H^*(D(E)/S(E)) = \tilde H^*(T(E)).$$

Note that to deduce this we would need the pair to be a good pair, but this will be easy for you to show (if not completely obvious by what you already did). You need to show that $S(E)\subset D(E)$ is nonempty and closed and that there exists a neighborhood which deformationretracts onto it.