Note, that a proof of this is already done on Parametrization of $n$-spheres. However, I wanted to approach the proof differently and was wondering if my line of logic would work.
Let $S^{n}$ be the $n$ sphere. Suppose it can be covered via one parameterization. Then, $\exists$ a mapping $\phi:U\subseteq\mathbb{R}^k \to S^n$ (for some open $U$ ).
However, note that $S^n$ is compact and spherically symmetric so that it attains two (distinct) points on $S^n$ which are the farthest from the origin such that only the last $n-k$ coordinates of each values are different. (here I tried to think of the sphere in three dimensions and the points at the north and south pole, they are essentially identical except for the last coordinate). However, I cannot seem to be able to prove the existence of such two points and am just hand-waving its existence. Can someone try to help me at this step? Of course, if such two points do not necessarily have to exist, then I would like to know that as well.
If I can prove this, then it would follow that $\phi$ is not injective so it cannot be a parameterization (I am taking the definition that parameterizations need to be diffeomorphic).
You are complicating what is, in essence, a simple problem. Remember that, in particular, diffeomorphisms are homeomorphisms. If $\emptyset \ne U = \mathring U \subseteq \Bbb R^k$ is diffeomorphic to $S^n$, then it is homeomorphic to it. Since $S^n$ is compact, $U$ too will have to be so. In particular, it must be closed. Since $U$ is also open and non-empty, and $\Bbb R^k$ is connected, $U$ must be the whole space $\Bbb R ^k$. We deduce that $\Bbb R^k$ is compact - which is not possible, since it is unbounded, therefore $S^n$ cannot be diffeomorphic to any open subset of $\Bbb R^k$.