TITLE: Prove the unit circle is uncountable
QUESTION [7 upvotes]: This is a homework problem, so avoid giving the answer. I think a discussion of my attempt at a proof would be more appropriate. The problem goes as follows:
Let $S$ be the circle of unit radius in the Euclidean plane:
$$S = \{ (x,y) \in \mathbb{R}^2:   x^{2} + y^{2}=1 \}$$
Prove that $S$ is uncountable. 
This is my attempt at a proof. I don't know if it is valid, or if my logic, and for that matter my approach to the proof, is correct. Feedback/comments/thoughts of any kind are welcome. 
Let $G^{+} = \{(x,y)\in G: y \geq0\}$ and $G^{-} = \{(x,y)\in G: y \leq 0 \}.$These are the upper and lower segments of the unit circle. 
Notice that $G^{+}\subset S$ and $ \hspace{1mm}G^{-}\subset S$, so $S=G^{+} \cup G^{-}.$ 
Let $f: G^{+} \rightarrow [-1,1],$ where $f(x,y)=x$. This can be thought of as a projection of the semi-circle onto the $x$-axis. 
Since the image of $G^{+}$ under $f$ is equal to the codomain; i.e., $f(G^{+})= [-1,1]$, then $f$ is surjective. 
Now since for every $(x,y) \in G^{+}$ we have the cardinality $|f(x,y)|=1$, there exists an inverse function $f^{-1}:[-1,1] \rightarrow G^{+}$ defined by $f^{-1}(x) = (x,\sqrt{1-x^{2})}$. 
Thus, there exists a bijection between $G^{+}$ and $[-1,1].$
Similarly we have the same argument for $G^{-}.$ Let $g:G^{-} \rightarrow [-1,1]$ then 

$ g(G^{-})=[-1,1]$ (surjective)
$|g(x,y)|=1 \hspace{4mm}  \forall (x,y)\in G^{-}$ (one-to-one)
$g^{-1}: [-1,1] \rightarrow G^{-}, \hspace{4mm} g^{-1}(x)= (x,-\sqrt{1-x^{2}})$ (inverse) 

which shows the bijection. 
Since the set of real numbers $[-1,1]$ is uncountable as can be shown by Cantor diagonalization, and we have $G^{+} {\raise.17ex\hbox{$\scriptstyle\sim$}} [-1,1]$ and  $G^{-} {\raise.17ex\hbox{$\scriptstyle\sim$}} [-1,1]$, then that implies that $G^{+}$ and $G^{-}$ are uncountable. Therefore $S= G^{+} \cup G^{-}$ must also be uncountable. q.e.d.
One other approach I thought about was to think of the semicircles as intervals in their own right, where the length of the upper semicircle would be $[0,\pi]$ and the interval of the lower semicircle would be $[\pi,2\pi],$ and I suppose the metric would be the arc length. So essentially you take the arc length and straighten it out, but I didn't know how to approach it or even formalize it. However, I think it is essentially the same thing as what I did in my proof.

REPLY [4 votes]: I think the proof is fine. 
You might also like to note that a unit circle can be charactherized by using the corresponding angle, $\theta \in [0, 2\pi)$.
Since $[0, 2\pi)$ is uncountable, you obtain your result as well.