Prove that $ρ$ induces a continuous function from $S^n → O(n+1)$ and therefore from $\Bbb RP^n → O(n+1)$ 
For $v∈S^n$ define $ρ_v$ to be the reflection along the subspace generated by $v$. Prove that $ρ$ induces a continuous function from $S^n → O(n+1)$ and therefore from $\Bbb RP^n → O(n+1)$.

My attempt:
My idea (not a proof, just an idea!) : Let us consider the special case $v = e_1$ , then consider $\{e_1\}^{\perp}$ in $S^n$ . $\{e_2,\dots,e_n\}$ is  a Orthonormal Basis of $\{e_1\}^{\perp}$ in $S^n$. Then according to definition of $\rho$, $\rho_{e_1} (e_j)=-e_j , \forall j \ge 2$, thus we get the map $\rho_{e_1}: S^n \to S^n$ defined by,$$ \rho_{e_1}(e_i)=\begin{cases} e_1 & i=1 \\ -e_i  & i \ne 1 \end{cases}$$ Clearly, $\rho_{e_1} \in O(n+1)$ .
(1)Thus, for each $v \in S^n$ , by fixing an orthonormal basis  of $\{v\}^{\perp}$, can we repeat the above argument?
(2) If the answer of the above Question is yes,  how to ensure Continuity of $v \mapsto \rho_{v}$ ? Does noting that $S^n$ is connected and hence it has only two orientations help?
(3) Does the answer for "map from $\Bbb RP^n → O(n+1)$" follows from the following diagram?

 A: A hyperplane $H  \subset \mathbb{R}^{n+1}$ is an $n$-dimensional subspace of $\mathbb{R}^{n+1}$. Each hyperplane $H$ determines the reflection $\rho_H$ in $H$. It should be geometrically clear what this means, but we need more precision.
The orthogonal complement $H^{\perp}$ of $H$ is a one-dimensional subspace of $\mathbb{R}^{n+1}$. Thus $H^{\perp} \cap S^n$ consists exactly of two points $v^\pm_H$ such that $v^-_H = -v^+_H$. We have $(v^\pm_H)^{\perp} = H$. Now each $w \in \mathbb{R}^{n+1}$ splits uniquely as
$$w = (w \cdot v^\pm_H) v^\pm_H + (w - (w \cdot v^\pm_H) v^\pm_H)$$
where the first summand $(w \cdot v^\pm_H) v^\pm_H$ is in $H^{\perp}$ and the second summand $w - (w \cdot v^\pm_H) v^\pm_H$ is in $H$. Then we get
$$\rho_H(w) = -(w \cdot v^\pm_H) v^\pm_H + (w - (w \cdot v^\pm_H) v^\pm_H) = w  - 2(w \cdot v^\pm_H) v^\pm_H .$$
Let us conversely start with $v \in S^n$. The orthgonal complement $v^{\perp}$ of $v$ is a hyperplane so that we get the reflection
$$\rho_v = \rho_{v^{\perp}}.$$
Explicitly we have
$$\rho_v(w) = w - 2(w \cdot v) v .$$
Note that $\rho_{-v} = \rho_v$. This shows that $\rho : S^n \to O(n+1)$ induces
$$\hat{\rho} :\mathbb{R}P^n \to O(n+1) .$$
It remains to show that $\rho$ is continuous. Let us regard it as a map into the vector space $E = End(\mathbb{R}^{n+1}) \supset O(n+1)$ of linear maps $\mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$. The space $E$ can be endowed with the operator norm $\lVert \phi \rVert_{op} = \sup \{ \lVert \phi(w) \rVert : \lVert w \rVert = 1 \}$. We have
$$\lVert \rho_v - \rho_u \rVert_{op} = \sup \{ \lVert - 2(w \cdot v) v + 2(w \cdot u) u \rVert : w \in S^n \}$$
$$= 2\sup \{ \lVert (w \cdot v) v - (w \cdot u) u \rVert : w \in S^n \} .$$
With $d = v - u$ (i.e. $v = u + d$) we easily compute
$$\lVert (w \cdot v) v - (w \cdot u) u \rVert = \lVert (w \cdot d) v + (w \cdot u) d \rVert \le \lVert (w \cdot d) v \rVert + \lVert (w \cdot u) d \rVert$$
$$= \lvert w\cdot d \rvert \lVert v \rVert +  \lvert w\cdot u \rvert \lVert d \rVert \le \lVert w \rVert \lVert d \rVert \lVert v \rVert +\lVert w \rVert \lVert u \rVert \lVert d \rVert  = 2\lVert d \rVert .$$
Recall that $u,v, w \in S^n$. This shows
$$\lVert \rho_v - \rho_u \rVert_{op} \le 4 \lVert v -  u \rVert .$$
Hence $\rho$ is continuous.
