Finding counterexamples: bijective continuous functions that are not homeomorphisms
Second edit: Your new examples are both correct.
First edit: I wrote that the first example was incorrect. That was a typo. It is actually correct.
Your first example is correct. You need to show that $f^{-1}$ is not continuous, that is, that it is not true that $(f^{-1})^{-1}(U)=f(U)$ is an open set for all open sets $U$. Take any set $U$ which is open in $X$ but which is not open in the indiscrete topology, for example $U=(1/3,1/2)$. Then $f(U)=U$, which is not open in $Y$. Thus $f$ is not a homeomorphism.
Your second example is incorrect because there is no bijective continuous $f$ from $(0,1)$ to $[0,1]$: you must have $f(x)=0$ for some $0<x<1$, so by the intermediate value theorem, there must exist $y<x<z$ such that $f(y)=f(z)=\epsilon$ for some small $\epsilon>0$ (think of the curve of $f$ dipping down to $0$ and then going up again; it must pass through equal values somewhere on opposite sides of the zero).
Let $f: X \to Y$ be the identity function, where $Y=X$. It is obvious a bijection. If you need $f$ is continuous mapping, not a homeomorphism. It only need the topology on $Y$, we say $\tau_Y$, is strictly weaker that the topology $\tau_X$ on $X$. Then $f$ must be continuous, however it is not open, hence it is not a homeomorphism.
Proof: It is continuous. Let $U \in \tau_Y \subseteq \tau_X$, then $f^{-1}(U)= U \in \tau_X$, which witnesses $f$ is continuous. It is not open. There exist $U \in \tau_X \setminus \tau_Y$, then $f(U)=U \notin \tau_Y$, which shows $f$ is not open.
Note Discrete topology is the strongest topology on the given set. And indiscrete topology is the weakest topology on the given set.
Hope it be helpful for you also.
If $X$ is not compact, then the theorem does not work. For example, take $\mathbb R$ with the discrete topology and let $f$ be the identity into $\mathbb R$ with the standard topology.
But Hausdorffness isn't really necessary. It suffices that compact subsets are closed. There exist compact spaces where the compact subsets are exactly the closed subsets, still, the space is not Hausdorff. These spaces are called maximal compact. The reason is that these are exactly the spaces where there is no finer topology on the space $X$ such that the space is still compact. It is relatively easy to show that, if the compact subsets are exactly the closed ones, then no finer topology leaves the space compact. The other direction is more difficult. For a proof and an example of such a space, see this paper.