Problem: If $x:N^{n}\to \mathbb{R}^{n+1}$ os umbilical, them $x(N)$ is contained in a n-plane or a n-sphere.
I know that a immersion is umbilical if $\langle B(X,Y),\eta\rangle=\lambda(p)\langle X,Y\rangle$, $\lambda(p) \in \mathbb{R}$, for all $p$ and if, the curvature of codomain is constant, then $\lambda$ is independent of $p$. Then, I thought to break the proof in two cases: $\lambda=0$ and $\lambda\ne 0$.
If $\lambda=0$,
$-\langle \nabla_{X} ,\eta\rangle=0$, but I don't know how to continue...
OBS: $X,Y$ are differentiable vector fields to $N$, $\eta$ is normal unitary vector field to $x(N)$ and $\langle\cdot,\cdot\rangle$ is a metric $g$ on Euclidean space and the metric induced in $N$ by $x$.
Thanks
Mimic the usual proof in the low-dimensional case. I'll write $M$ instead of your $N$, since I want to use $N$ for the normal field on "$M$".
Let $f\colon U \subseteq \Bbb R^n \to M \subseteq \Bbb R^{n+1}$ a parametrization of $M$ such that $U$ is connected. Putting $N(u) \equiv N(f(u))$ we have $$-{\rm d}N_{f(u)} = \lambda(u) {\rm Id}_{T_{f(u)}M}$$for some smooth function $\lambda \colon U \to \Bbb R$. Then applying both sides above in $\partial_if(u)$ we have $$-\partial_iN(u) = \lambda(u) \partial_if(u).$$Differentiating in the $u_j$ direction we get $$-\partial_j\partial_iN(u) = \partial_j\lambda(u) \partial_if(u) + \lambda(u)\partial_j\partial_if(u).$$Swapping $i$ and $j$ we get a similar relation, and subtracting we obtain $$\partial_j\lambda(u)\partial_if(u) - \partial_i\lambda(u)\partial_jf(u) = 0,$$and linear independence yields $\partial_i\lambda \equiv 0$ for all $i$, and so $\lambda$ is constant.
If $\lambda = 0$, then $\partial_iN \equiv 0$ and $N$ is constant - hence $f[U]$ is contained in some hyperplane.
If $\lambda \neq 0$, put $c\colon U \to \Bbb R^{n+1}$: $$c(u) = f(u) + \frac{1}{\lambda}N(u).$$Then $\partial_ic\equiv 0$ for all $i$ (by construction), and so $c$ is a constant - our center. So $$\|f(u) - c\| = \frac{1}{|\lambda|},$$and $f[U]$ is contained in a sphere centered in $c$ with radius $1/|\lambda|$.
If $M$ is connected, whichever case above holds, it'll be globally.