Let be $x\in S^n\subseteq \mathbb{R}^{n+1}$ and $0<\varepsilon<1$ and $u\in B^{n+1}_\varepsilon(x)$.
1) How can I prove that $\frac{u}{\lVert{u}\rVert}\in B^{n+1}_\epsilon(x)$?
2) If I want to say that for each $\lambda\in (a,b)$ I have $\lambda u\in B^{n+1}_\epsilon(x)$, how should I take $a$ and $b$ such that the interval $(a,b)$ is the largest possible?
Maybe I have succeded in showing the point 1) but I think there can be a smarter and shorter proof than mine.
Your claim 1) is in fact false. Here is a counter-example:
Let $n=1$, $x=(1,0)$ and $\varepsilon=\frac12$. Then $u:=(\frac34, \frac{\sqrt3}4)\in B^{n+1}_\varepsilon(x)$ since $u-x=(-\frac14,\frac{\sqrt3}4)$ and $\Vert u-x\Vert^2=\frac14$. But $\frac{u}{\Vert u\Vert}=\frac{(\frac34, \frac{\sqrt3}4)}{\frac{\sqrt3}{2}}=(\frac{\sqrt3}{2},\frac12)$ and $\Vert\frac{u}{\Vert u\Vert}-x\Vert^2=2-\sqrt3>\frac14$.