How to establish there is no continuous bijection? I know how to prove that there is a bijection between $(0,1)$ and $[0,1]$ but no continuous bijection between $(0,1)$ and $[0,1]$. 
What my confusion is the following: Suppose we need to establish that there is no continuous bijection between $(0,1)$ and $[0,1)$. I tried to proceed like this: Since $[0,1)$ is equipotent to $[0,1]$, hence searching continuous bijection between $(0,1)$ and $[0,1)$ is same as searching continuous bijection between $(0,1)$ and $[0,1]$. But the later has no such continuous bijection. Hence we are done.
Not sure if I made any mistake. Would some one correct me please ?
In general, will it be true that "For $A, B, C\subseteq \mathbb R$, if there is no continuous bijection between $A$ and $B$ where  $B$ is equipotent to $C$ then there will be no continuous bijection between $A$ and $C$," ?
 A: Note that equipotency only ask for a bijection to exist, there is no need for continuity, hence searching for a continuous bijection cannot be done by looking at equipotency. The other way round of course works: If $A$ and $B$ are not equipotent, there is no continuous bijection (as there isn't any bijection). 
If we need to establish that there is no continuous bijection between $(0,1)$ and $[0,1)$, we have to start all over, so suppose there were one, $f \colon (0,1) \to [0,1)$. Write $x := f^{-1}(0)$. Choose $\epsilon > 0$ such that $I := (x-\epsilon, x+ \epsilon) \subseteq (0,1)$. As $f$ is one-to-one and continuous, $f$ has to be monotone, but this is impossible on $I$, as $f$ has an inner minimum on $I$. Contradiction.
A: An alternative method to prove it:
One can use the fact that "Continuous image of a compact set is compact".
Here, $[0,1]$ is compact but $(0,1)$ is not.
A: There are several questions in your post.
Regarding the last one "For $A, B, C\subseteq \mathbb R$, if there is no continuous bijection between $A$ and $B$ where  $B$ is equipotent to $C$ then there will be no continuous bijection between $A$ and $C$," the answer is clearly negative, i.e. take $A=C=(0,1)$ and $B=[0,1]$.
A: Yet another alternative. If there existed a homeomorphism from [0,1] to (0,1), then you can use a connectness argument and you would find a contradiction.
The basic idea of this argument relies on the fact there are two points we can remove from [0,1] such that the result remains connected (since it is a path connected Euclidean space). However, any point removed from (0,1) causes the result to be disconnected.
But honestly, the compactness argument is all you really need!
