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Here $i$ is the inclusion map, $D$ denotes the disk and $S$ denotes the sphere. I would be interested in an answer using homology of pairs. My idea was to assume that $i$ is a homotopy equivalence of pairs, which implies that the $H_n(D^n,S^{n-1})$ is isomorphic to $H_n(D^n,D^n\setminus \{0\})$. Then by calculating each homology group separately I hope to get a contradiction.

Edit: The original idea was almost the right one. The Homology groups of the pairs are isomorphic. However, the homotopy equivalence of pairs also induces isomorphisms of slightly modified pairs, which then imply the contradiction. I provided a detailled answer.

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    Those are both isomorphic (and the inclusion induced an isomorphism.) Try to write down the inverse map in the equivalence and see what goes wrong with the inverse (the inclusionnis fine). – 2017-01-20
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    Just to clarify, it is possible to have isomorphic homology groups even if the corresponding spaces are not homotopy equivalent, but homotopy equivalence always induces isomorphisms between the homology groups? I get the feeling that this is exactly the point of this excercise. – 2017-01-20
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    No, indeed your map does induce an isomorphism on homology (and homotopy) groups. The point of this exercise is that a homotopy equivalence has an inverse (up to homotopy). When you pretend you have such an inverse, you will derive such a contradiction. – 2017-01-20

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Assume for contradiction that $i:(D^n, S^{n-1}) \to (D^n,D^n\setminus \{0\})$ is a homotopy equivalence of pairs. Then $i$ also defines a homotopy equialence of the pairs $(D^n, S^{n-1})=(D^n, \overline{S^{n-1}}) \to (D^n,\overline{D^n\setminus \{0\}})=(D^n,D^n),$ (see this question).
Thus $(D^n, S^{n-1})$ and $(D^n,D^n)$ have isomorphic homology groups.

But this is a contradiction:
$H_n(D^n,D^n)=0$: This follows directly from the definition of relative homology. Alternatively use the long exact sequence of the pair $(D_n,D_n)$

On the other hand, $H_n(D^n,S^{n-1})\neq 0$:
Either use that $H_n(D^n,S^{n-1})=H_n(D^n/S^{n-1})=H^n(S^n)\neq0$ or alternatively use the long exact sequnce of the pair $(D^n,S^{n-1})$ to conclude $H_n(D^n,S^{n-1})=H_{n-1}(S^{n-1})\neq 0$