This is an exercise from the book Espaços Métricos (metric spaces) by Elon Lima. I'm translating it (the part of it that I'm having trouble with):
Show that if $X\subset\mathbb R^n$ is such that $\left\|x-y\right\|=1$ whenever $x\neq y$ and $x,y\in X$, then $X$ has at most $n+1$ points.
This is pretty straightforward when $n=1$, easy with some algebra when $n=2$, a little bit exhaustive with lots of algebra when $n=3$, but I'm not able to generalize the argument to $\mathbb R^n$ at all.
I know this looks like induction, but I've been trying for some time, and I can't do it.
Let's try induction on $n$. Say you've proved it for $1,2,3,\ldots,n-1$. Suppose we have a set of $n$ points at unit distance from each other in $\mathbb R^n$ and we seek more points in that space that we can add to that set of $n$ points. Those $n$ points must be affinely independent and affinely span an $(n-1)$-dimensional affine subspace of $\mathbb R^n$, since if they spanned a space of lower dimension than that, then our proposition would fail to be true in at least of the cases $1,2,3,\ldots,n-1$. No generality is lost by assuming that that $(n-1)$-dimensional subspace is $\{(x_1,\ldots,x_{n-1},0) \in\mathbb R^n\}$. Any point in $\mathbb R^n$ at unit distance from all $n$ of those points in that affine subspace must be in either of two hyperplanes parallel to that subspace. The two points in those planes at unit distance from all $n$ of the points we started with are not at unit distance from each other, so we must discard one and keep one.