Induction proof that the $n$-sphere $S^{n}$ is connected

Question

I want to prove, by induction, that $S^n$ is connected.

1)First step of the induction:

$S^1$ is the image of $[0,1]$ by a continuous function, so it is connected.

2)Let's make the hypothesis that $S^{n-1}$ is connected.

For every point $x,y \in S^n \subset \mathbb{R^{n+1}}$, there is an hyperplane $H$ passing by $x,y,O$, where $O$ is the origin of $\mathbb{R^{n+1}}$. Then $x,y \in S^{n-1}\subset H $, for all $x,y \in S^{n}$, where $S^{n-1}$ is connected by the induction hypothesis. Since every pair of points of the set belong to a connected subspace, $S^{n}$ is connected.

Is this approach correct? I am not sure about the idea of the hyperplane in higher dimension, since it is not unique.

Thanks in advance, every help is appreciated.

Answer 1

Your idea works. For $n \ge 3$ hyperplanes in $\mathbb R^{n+1}$ are not uniquely determined by the three points $x,y, O$ but that does not matter. Simply let $H$ be any $n$-dimensional linear subspace containing $x, y$. You have a little gap here: You must show that $H \cap S^n$ is homeomorphic to $S^{n-1}$, but this is not hard.

Answer 2

If I really had to use induction, I would prove the connectedness of the sphere $S^n$ observing that $$ S^{n+1}\simeq \frac{S^n\times[-1,1]}\sim $$ where $(x,-1)\sim(y,-1)$ and $(x,1)\sim(y,1)$ for all $x$, $y\in S^n$ are the only non-trivial equivalences.

This allows to set up a quite simple induction since products of connected spaces and quotients of connected spaces are connected.