how many unit vectors are there in $\mathbb{R}^n$? How many vectors $\vec{x}$ in $\mathbb{R}^n$ ($n \geq 2$) satisfy the condition that $||\vec{x}||=1$?
First of all, it is obviously that one would say there are infinitely many such vectors in $\mathbb{R}^n$. However, I want a proof for this claim. 
I asked two of my professors. Both of them gave me the same answer: There are infinitely many such vectors. Because if we consider the $\mathbb{R}^2$ plane and draw a unit circle. There are infinitely many points in the circumference since each angle corresponds to one point and we obviously have infinitely many angles in $\mathbb{R^2}$.
I understand that my professors were trying to use a simple illustration to help me think about it intuitively. But I want a mathematical proof of these statement:
(1) How can you prove that the circumference of a circle contains infinitely many point?
(2) How can you prove that there are infinitely many angles and each angle correspond to one point?
Again, I know these are obviously true statements, but we can not prove a statement by simply saying that it is obvious true.
Is it  because of axioms of geometry or definition of $\mathbb{R}^2$ or conventions made by mathematicians so that the above statements are true?
 A: There certainly is philosophical content to this question. And we can respond in a fundamental way without using any alleged properties of "real numbers"... by considering vectors $({2n\over n^2+1},{n^2-1\over n^2+1},0,0,\ldots)$ for positive integers $n$.
EDIT: (in light of comments...) For what it's worth, again, this recipe does not use trig functions, and does not use real numbers. On another hand, we can let $n$ range through positive real numbers (or some avatar thereof), if desired. It all depends on what in what context one wants the question answered, obviously. But this "Pythagorean triple" formula is pretty "close to the metal"...
A: For $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$,
$$\vec x = \ \big(\sin \theta , \cos \theta , \underbrace{0 , \dots , 0}_{n-2 \text{ times}}\big) \in \mathbb R^n$$
is a unit vector in $\Bbb R^n$. Now since $\sin \theta$ is injective on the given domain, for different $\theta$, we find different unit vectors. There are infinitely many $\theta$ values within the given interval. The result follows.
A: Just consider the upper-right quadrant of the unit circle. 
I presume you are happy to admit there are infinitely many points in the interval $[0,1]$.
Each $x \in [0,1]$ is the $x$-coordinate of a point on the upper-right quadrant. 
The $y$-value of that point is is $y= \displaystyle \sqrt{1-x^2}$ by Pythagoras' theorem.
So each of the infinitely-many $x \in [0,1]$ is the $x$-coordinate of a unit vector. 
These vectors are all distinct because their $x$-coordinates are all distinct.
But we can be more explicit! 
We can in fact write down infinitely many allowed $x$-values.$^1$ For example the sequence $1, \frac{1}{2}, \frac{1}{3}, \ldots $ of infinitely many real numbers. 
The same method as before lets us write down infinitely many different unit vectors:
$\displaystyle \Big ( \frac{1}{2} , \sqrt{1-\frac{1}{4} } \Big )$, 
$\displaystyle \Big ( \frac{1}{3} , \sqrt{1-\frac{1}{9} } \Big )$, 
$\displaystyle \Big ( \frac{1}{4} , \sqrt{1-\frac{1}{16} } \Big ), \ldots$
And you can check they are unit length by taking inner prododucts.

$1$. If you want to show there are uncountably many real numbers, we cannot write them all down. And you need to specify what you ARE assuming about the real number line to start.
A: If we can simply exhibit infinitely many unit vectors in $\Bbb R^2$, we will be done because if $n\ge 2$ we can find a copy of $\Bbb R^2$ inside of  $\Bbb R^n$ via the map $(x,y)\mapsto(x,y,0,...,0)$.
Consider the points $p_n=(\cos\frac{1}{n},\sin\frac{1}{n})$, where $n = 1,2,3,\dots.$ This is an infinite set because all the points $p_n$ are distinct, i.e., $p_n\ne p_m$ if $n\ne m$. To justify this, note that $\cos$ is injective on $[0,1]$ (draw the graph), and all of our points $p_n$ lie in $[0,1]$. Thus, the value of $\cos\frac{1}{n}$ never repeats itself as $n$ changes, hence the points $p_n$ are all distinct.
The usual trig identity $\cos^2\theta+\sin^2\theta=1$ shows us that each point $p_n$ is a unit vector. Hence there are infinitely many unit vectors in $\Bbb R^2$, and in $\Bbb R^n$, as desired.
