Equivalence of definitions of $S^\infty$

Solution 1:

For $n<m$ we can isometrically embed $\mathbb R^n\to \mathbb R^m$ by mapping to the first coordinates / filling up with zeroes. This way the limit of all $\mathbb R^n$ is $\mathbb R^\infty$, the set of all sequences $(x_n)_{n\in\mathbb N}$ with almost all terms $=0$, which still gets its topology from the scalar product $\langle x,y\rangle=\sum_{i=1}^\infty x_iy_i$ (that works because the sum is actually finite).

Consider the set $S^\infty:=\{x\in\mathbb R ^\infty\mid \lVert x\rVert = 1\}$. For each $n\in\mathbb N$ and $\epsilon\in \{\pm1\}$ we have the subset $$C_{n,\epsilon}=\{x\in S^\infty\mid x_n\epsilon>0\land \forall k>n\colon x_k=0\}.$$ Then $C_{n,\epsilon}$ is just one hemisphere of $S^{n-1}\subset \mathbb R^n$ and the smaller ones make up the equator. More precisely, the map $$(x_1,\ldots,x_{n-1})\mapsto (x_1,\ldots, x_{n-1},\epsilon \cdot\sqrt{1-x_1^2-\ldots-x_{n-1}^2}, 0,0, \ldots) $$ defined on the closed $(n-1)$-ball is a homeomorphism of its interior with $C_{n,\epsilon}$ and of its boundary with the union of all smaller $C_{n',\epsilon'}$ and at the same time $S^{n-1}\subset \mathbb R^{n-1}\subset \mathbb R^\infty$.

This shows the CW-structure you describe in "2)".