$D^2$ is not homeomorphic to $D^n/S^{n-1} $ The following example was left in exercise of my topology class. I think I would need help on how to prove the asserstion asked.
Let $D^n$ denote the unit ball in the n-th dimensional euclidean space and let $S^{n-1} $be the unit sphere.

Show that $D^2 $ is not homeomorphic to $D^n /S^{n-1}$.

These types of questions are proved by assuming that there exists an homeomorphism and then there is a fundamental property( upto homeomorphism) which is not satisfied in one of the sets but is satisfied in the other sets.

But I am not able to think of such a property in the current question.


As Randall already pointed out in the comments, $D^n/S^{n-1}$ is homeomorphic to a sphere, in this case $S^n$. This is proven here by Henno Brandsma using the one-point or Alexandroff compactification (See here). In short, the argument looks like this:
$$D^n/S^{n-1}
\cong(D^n\setminus S^{n-1})^*
\cong((D^n)^\circ)^*
\cong(\mathbb{R}^n)^*
\cong S^n.$$
The first step is the fifth point here, the second step is using $\partial D^n=S^{n-1}$, the third step is using $(D^n)^\circ\cong\mathbb{R}^n$ (See here.) and the forth step can be imagined as stereographic projection and adding the missing point.
To now solve your problem: $D^2$ and $D^n/S^{n-1}\cong S^n$ are not homeomorphic (and not even homotopy equivalent) since their $n$-th homotopy groups are not isomorphic:
$$\pi_n(D^2)
=1\not\cong\mathbb{Z}
\cong\pi_n(S^n)
\cong\pi_n(D^n/S^{n-1}).$$
You could also say, that $D^2$ is contractible (homotopy equivalent to a point, meaning all homotopy groups are trivial), while $D^n/S^{n-1}$ is not.
