On existence of subset of $\mathbb{R}^{n}$ homemorphic to $\mathbb{S}^n$

If by contradiction you could find $S\subset \Bbb R^n$ and a homeomorphism $\phi:S\to \Bbb S^n$, then the composition of $\phi^{-1}$ with $i:S\to \Bbb R^n$ would give a continuous injection $\Bbb S^n\to \Bbb R^n$. This is in contradiction with the Borsuk–Ulam theorem, which says in particular that any continuous map $\Bbb S^n\to \Bbb R^n$ can't be injective.

Here's a generalization: if $M$ is any compact $n$-manifold then there is no subset of $\mathbb R^n$ homeomorphic to $M$. For suppose there were, and let $f : M \to \mathbb R^n$ be a homeomorphism onto its image $f(M)$. By the invariance of domain theorem, $f(M)$ is an open subset of $\mathbb R^n$. Since $f(M)$ is compact, it is also a closed subset of $\mathbb R^n$. Since $\mathbb R^n$ is connected, it follows that $f(M)=\mathbb R^n$, but $\mathbb R^n$ is not compact.