Prove that $R^{n}\setminus R^{k} \simeq S^{n} \setminus S^{k} \simeq S^{n-k-1} $ $\simeq$ is homotopy equivalence space
homotopy equivalence: 
Two topological spaces X and Y are homotopy equivalent if there exist continuous maps $f:X\rightarrow Y$ and $g:Y \rightarrow X$, such that the composition f degreesg is homotopic to the identity $id_Y$ on Y, and such that g degrees f is homotopic to $id_X$. Each of the maps f and g is called a homotopy equivalence, and g is said to be a homotopy inverse to f (and vice versa).
homotopy: 
In topology, two continuous functions from one topological space to another are called homotopic if one can be "continuously deformed" into the other, such a deformation being called a homotopy between the two functions.
$\setminus$ is set minus (dont mistake it with quotient space)
hint: there is similar question here but its not usable for this question because first of all it proves $R^{n}\setminus R^{k} \simeq S^{n-k-1} \times R^{k+1}$
The answer tried to solved it by using induction on k,
 and the right term  $R^{k+1}$ play important roll which can not be omitted from that proof. and if I wanted to prove it by induction on k 
k=0
$R^{n} \setminus R^{0} \simeq S^{n-1}$
I can prove the initial case very similar by giving the map: $x \rightarrow (x/ ||x||)$
however I have no idea for the induction step. and for the middle part of the proof which says: $\simeq S^{n} \setminus S^{k}$
 A: Your question is imprecise because you do not specify how $\mathbb R^k$ is regarded a subspace of $\mathbb R^n$ (similarly for $S^k$ and $S^n$). But certainly you identity 
$\mathbb R^k$ with $\{(x_1,\ldots,x_n) \in \mathbb R^n \mid x_{k+1} = \ldots = x_n = 0 \}$ and $S^k$ with $\{(x_1,\ldots,x_{n+1}) \in S^n \mid x_{k+2} = \ldots = x_{n+1} = 0 \}$.
Let $p = (1,0,\ldots,0) \in S^n$. Stereographic projection gives us a homeomorphism
$$h : S^n \setminus \{p \} \to \mathbb R^n, h(x_1,\ldots,x_{n+1}) =\left(\frac{x_2}{1-x_1},\ldots,\frac{x_{n+1}}{1-x_1}\right) .$$
We have $h(S^k \setminus \{p \}) = \mathbb R^k$, thus $h((S^n \setminus \{p \}) \setminus (S^k \setminus \{p \})) = \mathbb R^n \setminus \mathbb R^k$. But clearly $(S^n \setminus \{p \}) \setminus (S^k \setminus \{p \}) = S^n \setminus S^k$ which proves that $\mathbb R^n \setminus \mathbb R^k$ and $S^n \setminus S^k$ are homeomorphic.
As you stated in your question, $\mathbb R^n \setminus \mathbb R^k$  and $S^{n-k-1} \times \mathbb R^{k+1}$ are homotopy equivalent which implies $\mathbb R^n \setminus \mathbb R^k \simeq S^{n-k-1}$ because $\mathbb R^{k+1}$ is contractible.
