Proof: Show there is set of $n+1$ points in $\mathbb{R}^n$ such that distance between any two distinct points is $1$?

This approach is most reminiscent of your last question. In $\mathbb R^{n+1},$ the standard unit axis vectors such as $(1,0,0,...,0)$ are all $\sqrt 2$ apart. So, multiply by $(n+1),$ all the vectors $(n+1,0,0,...0)$ are $(n+1) \sqrt 2$ apart. These lie in the hyperplane where the sum of the coordinates is $(n+1).$ This hyperplane can be translated at then rotated so that it coincides with $\mathbb R^n,$ then shrunk by a common scalar factor so the distances are $1.$

Let me do the translation anyway: subtract $1$ from all coordinates, so the $n+1$ entries of the first vector are $(n,-1,-1,...,-1)$ and the sum is now $0.$ A normal vector to this hyperplane is $(1,1,1,...,1).$ If you can figure out how to rotate that into $(0,0,0,...,\sqrt {n+1})$ you will hhave successfully rotated all the points into the plane $x_{n+1}=0,$ which is $\mathbb R^n.$

Hint: For the first part, you can find an explicit example for any $n$. First put $n$ standard-basis unit vectors, and note the pairwise distance between any two unit vectors is the same (you will need to normalize the vectors so that the distance is $1$). Now, carefully choose the $(n+1)$th (left to you as exercise) vector.