Define $F : \mathbb{S}^n \times I \to \overline{\mathbb{B}^{n+1}}$ by $F(x, s) = sx$, show that $F$ is injective 
Define $F : \mathbb{S}^n \times I \to \overline{\mathbb{B}^{n+1}}$ by $F(x, s) = sx$, show that $F$ is injective on $\overline{\mathbb{B}^{n+1}} \setminus \{0\}$

I tried to show that by contradiction but it eventually went from being a set theoretic issue to an arithmetic issue to resolve, and I'm not sure how to conclude the proof
My Attempted Proof: 
Suppose there exists an $\alpha \in \overline{\mathbb{B}^{n+1}} \setminus \{0\}$ for which $F(x_1, s_1) = s_1x_1 = s_2x_2 = F(x_2, s_2) = \alpha$ for $(x_1, s_1) \neq (x_2, s_2) \in \mathbb{S}^n \times (0, 1]$. Then $x_1 = (x_{1}^{1}, ..., x_{1}^{n+1})$ and $x_2 = (x_{2}^1, ..., x_{2}^{n+1})$, so $s_1x_1 =s_2x_2$ implies $$s_1x_{1}^{i} = s_2x_{2}^{i}$$
for $s_1, s_2 \in (0, 1]$ and $d(x_1, 0) = 1$ and $d(x_2, 0) = 1$. Note that the superscripts on are usually sub-subscripts but I've just put them as superscripts so they can be seen better.
Now how can I show that for any choice of $s_1$ and $x_1$ subject to the two constraints $s_1 \in (0, 1]$ and $d(x_1, 0) = 1$ and for any choice $s_2$ and $x_2$ subject to the two constraints $s_2 \in (0, 1]$ and $d(x_2, 0) = 1$, that for each $i \in \{1, ... ,n+1\}$ we have $$s_1x_{1}^{i} = s_2x_{2}^{i} \implies s_1 = s_2 \ \text{and } x_1^i = x_2^i$$
This really seems to be an arithmetic issue, but I'm not sure how to show it rigorously. I'm guessing that it is something really trivial and something that I should know but for some reason I can't show it. If I can show the above, then I can show that $(x_1, s_1) = (x_2, s_2)$ contradicting the fact that $(x_1, s_1) \neq (x_2, s_2)$ and thus proving injectivity of $F$ on $\overline{\mathbb{B}^{n+1}} \setminus \{0\}$.

Suppose $F(s_1,x_1)=F(s_2,x_2)$, so that $s_1x_1=s_2x_2$. Taking the magnitude of both of these vectors, we get $s_1\|x_1\|=s_2\|x_2\|$, which since $x_1,x_2\in \mathbb S^n$ implies $s_1=s_2$. Therefore, $s_1x_1=s_1x_2$. Finally, since $s_1\neq0$ (which we know since $s_1x_1\in \mathbb B_{n+1}\setminus\{0\}$), we can scale both sides by $1/s_1$ to get $x_1=x_2$, proving injectivity.
