Showing that $f:(0,2\pi]\to S^1$ is not a homeomorphism 
I want to show that $f:(0,2\pi]\to S^1$ defined by $t\to (\sin t, \cos t)$ is not a homeomorphism. I will do this by showing that its inverse is not continuous. 

The inverse is defined by $(\sin t, \cos t)\to t$. So at $(0,1)$ we want to have $\lvert 2(1-\cos t)\rvert <\delta \Rightarrow \lvert t-2\pi\rvert<\epsilon$ for any $\epsilon$. If we take $\epsilon=\pi/2$, no matter how close $\cos t$ and $1$ are there will always be $t$'s close to $0$. So $f^{-1}$ is not continuous.
What do you think? Are there any mistakes? 
 A: Your proof is correct, but you could have saved yourself some work:
$S^1$ is compact, $(0,2\pi]$ is not. Therefore they can impossibly be homeomorphic.
This doesn't only show that $f$ is not a homeomorphism, but also that no other map $(0,2\pi]\rightarrow S^1$ can ever be a homeomorphism.
Note: To elaborate a bit on that remark:
If $A$ is compact and $B$ is not compact, then there cannot exist a homeomorphism $f:B\rightarrow A$, because the inverse $f^{-1}$ would have to map the non-compact space $B$ onto the compact space $A$. That is impossible, so $f^{-1}$ cannot be continuous.
A: Intuitively, as $(\cos t,\sin t)$ goes past the point $(1,0)$ on the circle, then $t$ jumps abruptly from $0$ to $2\pi$ or vice-versa, so you have a jump discontinuity.
Your $\varepsilon$-$\delta$ proof is right.
One may also say this: under a continuous function (in this case $f^{-1}$), the inverse-image of every open set is open.  With this function, the inverse image of the open set
$$
\left\{ (\cos t,\sin t) : 0\le t<\frac{1}{10} \right\}
$$
is a subset of the circle that is not open.
A: $S^1$ cannot be homeomorphic to any subset of the real line. Indeed, $S^1$ minus one point is connected, while any connected (interval) of $\mathbb{R}$ minus one point of its interior is not connected.
