Being compact is necessary for a continuous bijection to have a continuous inverse 
Theorem: Suppose that $f:X \rightarrow Y$ is one-to-one, surjective and continuous. If $X$ is compact, Then $f^{-1}:Y \rightarrow X$ is also continuous. 

The proof for this theorem is pretty easy. We can show that the inverse of every close set in $X$ is closed in $Y$ (Am I right?)
Now I want to show that being compact is necessary. My example is:
$$f(t) = (\cos(t),\sin(t)), \forall t\in [0,2\pi)$$
$f$ is one-to-one, surjective and continuous. But $f^{-1}$ is not continuous.
Now I'm looking for more examples. Or maybe a set of infinite examples. 
 A: First, the theorem you state actually requires $Y$ to be Hausdorff as well. See Proofwiki for the proof in that case, and Alex Provost's comment for a counterexample if $Y$ is not Hausdorff.
What about examples where compactness is neessary?
Examples of this exactly correspond to examples of topologies $(X, \tau)$ and $(X, \tau')$ such that $\tau \subsetneq \tau'$.
Why? Since $f$ is a bijection between $X$ and $Y$, we may simply identify the points of $X$ and the points of $Y$. Then, the fact that $f$ is continuous says that every open set in $Y$ is an open set in $X$; to make its inverse not continuous, we simply require $X$ to have open sets that are not in $Y$. So $X$ has some topology, $\tau'$, and it must strictly contain $Y$'s topology $\tau$ ("contain" meaning under the identification given by the bijection).
If we also want the counterexample to have $\tau, \tau'$ Hausdorff, we should just ensure that $\tau$ is Hausdorff because any finer topology of a Hausdorff topology is Hausdorff.
So:


*

*Hagen von Eitzen's example is pretty good to start: take $\tau$ to be any topology you like, say the standard topology on $[0,1]$, and take $\tau'$ to be the discrete topology on that set.

*Your example with $\sin(t), \cos(t)$ is, in disguise, the following example: Take $X = [0,1)$. Let $\tau'$ be the subspace topology on $X$ inhereted from $[0,1]$. On the other hand, let $\tau$ be the topology on $X$ from the quotient space $[0,1] / \{0,1\}$. Then $\tau \subsetneq \tau'$ as we wanted.

*The weak topology $\tau$ on a (say, normed space) $X$ is strictly contained in the strong topology $\tau'$.
I'm sure there are many more examples.
