How would I say in words, the description of this set in mathematics notation? $$S^{n-1}=\{x\in\mathbb{R}^n:\|x\|=1\}$$
Would it be right to say, $S$ is the set of all vectors in $\mathbb{R}^{n-1}$ such that the norm of the vectors, by whichever specified notion of distance, is equal to one?
Why is $S^{n-1}$ used instead of $S^n$? If it is the set of all vectors of norm one in $\mathbb{R}^n$, then shouldn't it also have the same dimensions as $\mathbb{R}^n$? 
Like, my problem is, I don't see how I can approach the question I'm working on without being able to visualise such a set in space.
 A: For your first question, it is written in a strange way. For instance, there is no use for "by whichever specified notion of distance". The norm has already been specified. You could have only said:
"...is the set of the vectors in $\mathbb{R}^n$ which have norm equal to $1$"
Now, for your second question,
$S^{n}$ has an intrinsic nature. It can bee seen in many different topological ways, aside from being an embedded manifold inside $\mathbb{R}^{n+1}$. It is the one-point compactification of $\mathbb{R}^n$, the quotient of the disk by identifying the boundary, the smash product of $S^1 $ $n$ times etc. The invariant thing is: it is always a $n$-manifold. It is therefore more natural to assign the number $n$ in the notation.
A: Say instead,

$S^{n-1}$ is the set of all vectors in $\mathbb R^{n}$ such that the norm of the vectors, by whichever specified notion of distance, is equal to one.

(notice the subtle differences?) We call it $S^{n-1}$ because each point therein is contained in an open set which "looks like" (is homeomorphic to, if you are familiar) an open set in $\mathbb R^{n-1}$. In other words, $S^{n-1}$ has $n-1$ dimensions locally, not $n$. Of course, you could replace $n$ with $n+1$, and you would have a perfectly good definition of $S^n$ in terms of $\mathbb R^{n+1}$.
Your argument that as a subset of $\mathbb R^n$, the sphere ought to have dimension $n$ breaks down when we consider $$\{(x_1,...,x_n)\in\mathbb R^n:x_1=0\}.$$ This set is the same topologically as $\mathbb R^{n-1}$, so it has $n-1$ dimensions. For $S^{n-1}$, we are doing something similar.
Manifold families with an $n$ attached ($\mathbb T^n,\mathbb{RP^n},\mathbb{R}^n$,S^n etc...) are usually locally $n$ dimensional, so this index helps us understand the structure of the manifold.
Visualizing $S^{n}$ is not too hard when $n=1$ or $n=2$. These are just the circle and sphere sitting inside $\mathbb R^2$ and $\mathbb R^3$ respectively. In higher dimensions, it is harder to visualize, so the topologist has to rely on theorems and poorly drawn pictures to do his work. 
A: $S^{n-1}$ the surface of the unit hyper-sphere of the $n$-dimension real (Cartesian-coordinate) space.
The index is $n-1$ because it is a surface; thus having one less dimension than the space, $\Bbb R^n$ , in which that hyper-sphere resides.
A: The set you gave is the set of all vectors in $\mathbb{R}^n$ with norm one. Not the set of all vectors in $\mathbb{R}^{n-1}$ with norm 1.
Why do we call this set $S^{n-1}$? Well, if you study topology/manifolds, you'll see that $S^{n-1}$ is an $n-1$ dimensional manifold. Put more simply, while $S^{n-1}$ is a subset of $n$ dimensional real space, it locally looks like it has $n-1$ dimensions. So we label it like that.
It might be helpful to think of the case of $n=2$. Then $\mathbb{R}^n$ is just the plane, and $S^{n-1}$ is the set of all points of the plane with norm 1: the unit circle. The circle is one-dimensional (it's locally homeomorphic to a line segment, and any point on it can be described by just a single coordinate $\theta$). Similarly, $S^2$ is just the surface of the unit sphere in three dimensional space.
