Infinite sphere minus one point homeomorphic to the hyperplane of equation $x_1=0.$ Let $\mathbb{R}^n$ equipped with the Euclidean topology. Identifying the point $(x_1,\ldots,x_n)$ to $(x_1,\ldots,x_n,0)$ of $R^{n+1}$, we can define the "direct limit" $\mathbb{R}^\infty$ equipped with the direct limit topology i.e. $U \subseteq \mathbb{R}^\infty$ is open iff $U \cap \mathbb{R}^n \subseteq \mathbb{R}^n$ is open.
Denote $(b_n)_{k\in \Bbb{N}}$ to be the canonical basis of $\mathbb{R}^\infty$ so that the canonical basis of $\mathbb{R}^n$ is $(b_1,\ldots,b_n$ which can be written as $\sum_{k=1}^n x_nb_k$. 
Now define $S^\infty \subseteq \mathbb{R}^\infty$ to be the subset consisting of tuples such that $\sum_i |x_i|^2 = 1$, equipped with the subspace topology and $H^{\infty}$ be the hyperplane of equation $x_1=0.$
I would like to prove that $S^{\infty}\setminus \{b_1\}$ is homeomorphic to $H^{\infty}$.
I was thinking about $f: (x_2,\ldots,x_n)\mapsto (0,x_2,\ldots,x_n)$. But I think I misunderstood the question.
EDIT: Perhaps we can prove that given a point $P$ different from $b_0$ there exist a point $Q$ in $H^{\infty}$ such that $P,Q$ and $b_1$ are aligned and using it?
 A: Ok, so lets start with a more "handy" definition of $\mathbb{R}^{\infty}$:
$$\mathbb{R}^{\infty}=\big\{(x_n)_{n=1}^{\infty}\ \big|\ x_i\in\mathbb{R},\ \exists_{N}\forall_{j>N}\ x_i=0\big\}$$
So it's a set of all real sequences that are constantly equal to $0$ at some point.
Now we can look at $\mathbb{R}^n$ as a subset of $\mathbb{R}^{\infty}$ defined by
$$\mathbb{R}^{n}=\big\{(x_n)_{n=1}^{\infty}\in\mathbb{R}^{\infty}\ \big|\ \forall_{j>n}\ x_j=0\}$$
So unlike $\mathbb{R}^{\infty}$ in $\mathbb{R}^n$ we fix the index (at $n$) starting from which a sequence has to be constantly equal to $0$.
We define topology on $\mathbb{R}^{\infty}$ by setting: $U$ is open iff $U\cap\mathbb{R}^n$ is open in $\mathbb{R}^n$ for all $n$.
All of this is consistent with your approach but formally allows us to write:
$$\mathbb{R}^1\subset \mathbb{R}^2\subset \mathbb{R}^3\subset\cdots$$
$$\mathbb{R}^\infty=\bigcup_{n=1}^{\infty}\mathbb{R}^n$$

If we write $\mathbb{S}^n$ for a unit sphere in $\mathbb{R}^n$ then it is easy to see that
$$\mathbb{S}^1\subset \mathbb{S}^2\subset \mathbb{S}^3\subset\cdots$$
and we can define
$$\mathbb{S}^{\infty}=\bigcup_{n=1}^{\infty}\mathbb{S}^n$$
which is consistent with your definition of $\mathbb{S}^{\infty}$.
Now pick a point $v\in\mathbb{S}^{\infty}$ (note that it doesn't matter if it is a part of some basis or not). Obviously there exists $N\in\mathbb{N}$ such that $v\in\mathbb{S}^{n}$ for all $n>N$. In particular
$$\mathbb{S}^{\infty}\backslash\{v\}=\bigcup_{n=N+1}^{\infty}\mathbb{S}^n\backslash\{v\}$$
Note that the indexing on the bottom starts from $N+1$. I will implicitely assume that $n>N$ from now on.
We are almost ready. Now take $\varphi_n:\mathbb{S}^n\backslash\{v\}\to\mathbb{R}^n$ to be the $n$-dimensional sterographic projection . It is well known that $\varphi_n$ are homeomorphisms and are compatible with each other in the following sense: if $n<m$ and $u\in\mathbb{S}^n\backslash\{v\}$ then $\varphi_n(u)=\varphi_m(u)$ (note that $\mathbb{R}^n\subset\mathbb{R}^m$). Actually $\varphi_n$ can be chosen arbitrarly as long as they are homeomorphisms that satisfy the compatibility property.
The compatibility property makes the following function well defined:
$$\Phi:\mathbb{S}^{\infty}\backslash\{v\}\to\mathbb{R}^{\infty}$$
$$\Phi(u)=\varphi_n(u)$$
where $n$ is such that $u\in\mathbb{S}^n$. Obviously the definition does not depend on $n$. Furthermore the continuity of $\Phi$ follows from the definition of inductive topology. Finally $\Phi^{-1}$ is given by
$$\Phi^{-1}(u)=\varphi_n^{-1}(u)$$
This function is again well defined and continous by the same argument (note that $\varphi_n^{-1}$ are also compatible on $\mathbb{R}^n$). $\Box$
