Proving $\varphi^{-1}\colon S\to [0,1)$, is not continuous

Question

Do you guys agree with my proof? Thank you!

$\def\R{{\mathbb R}}$

Let $\varphi\colon [0,1)\to S$, be defined by $\varphi(t) = (\cos(2\pi t), \sin(2\pi t))$, where $S$ is the unit circle in $\R^2$, that is $S=\{(x_1,x_2)\colon x^2_1+x^2_2=1\}$. I will assume without proving that $\varphi\colon [0,1)\to S$ is bijective, and continuous.
I want to prove $\varphi^{-1}\colon S\to [0,1)$, the inverse of $\varphi$, is not continuous.

$\textbf{solution}$ By compactness of image set we know the continuous image of a compact set is compact. If $\varphi^{-1}$ were continuous, then $S$ being a compact set, its image set $[0,1)$, as $\varphi^{-1}$ is homeomorphic, would be a compact one but it is not.

Answer 1

Actually, I think you only need the end of the argument. Since you know that $\varphi:[0,1)\to S^1$ is continuous and bijective, if its inverse were continuous, we would have $[0,1)\cong S^1$. However, this is a contradiction for a number of reasons - among them the fact that one space is compact while the other is not.