Can a subset of $\mathbb{R}^n$ be homotopy equivalent to $S^n$? Can a subset of $\mathbb{R}^n$ be homotopy equivalent to $S^n$? I am pretty certain the answer is no and I suspect it might be provable using homology groups, but I do not see how. Note that, because $S^n$ is the one-point compactification of $\mathbb{R}^n$, this question is equivalent to asking whether $S^n$ is homotopy equivalent to one of its proper subsets.
Edit: By freakish's comment, the answer is no when the subset is a manifold. However, I am now less certain whether the answer is no for any subset by the example Tyrone showed.
Edit 2: As Maxime Ramzi pointed out, the answer is also no if $X$ is compact by Alexander duality.
 A: By a corollary of the generalized Poincaré duality, referred as Poincaré Alexander Lefschetz duality in the book Geometry and Topology of Bredon, you can derive the following precious tool :

Corollary 8.5 (Bredon - G. & T.) If $L$ is a proper compact subset of an orientable connected $n$-manifold $M$, then $\check H^n(L; G) = 0$ for any coefficient group $G$.

So, if $X$ is a compact subset of $\mathbb R^n$, $X$ can't have the (weak) homotopy type of an $n$-sphere. Now more generally, if $X\subset \mathbb R^n$ is any subset and if it is homotopy equivalent to the $n$-sphere, then you get maps $\mathbb S^n\to X$ and $X \to \mathbb S^n$ such that the compositions are homotopic to the identities. If $L$ is the image of $\mathbb S^n\to X$, then $\mathbb S^n\to L\to \mathbb S^n$ is homotopic to $1_{\mathbb S^n}$ so the maps induced in Cech cohomology verify that $$\check H^n(\mathbb S^n)\to \check H^n(L)\to \check H^n(\mathbb S^n)$$
is the identity, which is not possible because Corollary 8.5 of Bredon applies to $L$ which is compact and so, $\check H^n(L)=0$.
