Show that no two of the spaces $(0, 1), (0, 1], [0, 1]$ are homeomorphic 
Show that no two of the spaces $(0, 1),  (0, 1], [0, 1]$ are homeomorphic

My Attempted Proof
We have $(0, 1) \subset (0, 1] \subset [0, 1]$.
We also have $(0, 1] = (0, 1) \cup \{1\}$ and thus $|(0, 1)| \neq |(0, 1]|$ and thus no bijective mapping $f : (0, 1) \to (0, 1]$ exists, hence $(0,1)$ and $(0, 1]$ are not homeomorphic regardless of the topology defined on them.
The proof for the other cases are similar. $\ \square$

Is my proof correct? I've seen proofs which form a contradiction through connectedness. Is there any error in my arguments?
 A: 
We also have $(0, 1] = (0, 1) \cup \{1\}$ and thus $|(0, 1)| \neq |(0, 1]|$ and thus no bijective mapping $f : (0, 1) \to (0, 1]$ exists

False. There exists a bijective mapping from $(0,1]$ to $(0,1)$. For example, one of them is
$$f(x) = \begin{cases}\frac x2 & \text{if } x=\frac{1}{2^k}\text{ for some }k\in \mathbb N\cup\{0\}\\
x & \text{else}\end{cases}$$
which maps $1$ to $\frac12$, $\frac12$ to $\frac14$, $\frac14$ to $\frac18$ and so on, while leaving all the other numbers in place.
Your argument is flawed, because if a set is infinite, adding one more element does not increase its size. For example, the set of even integers is equipotent to the set of all integers.

To actually prove your statement, think about the open sets that contain $1$ in $(0,1]$.

Also, and please don't take this the wrong way, but you may be rushing in your studies a little. If you don't yet know that $(0,1)$ has just as many elements as $(0,1]$, then your knowledge of set theory is at quite a basic level (i.e., it's something you should learn in your first month or so of graduate level math), while the topic you are addressing is basic topology, which is slightly more advanced (i.e., typically second-year undergrad). I advise you to slow down and take it step by step (if you are self-learning) or grab some first-year books and go back to step one (if you are a student).
A: Use the commonly used method of removing points from your space. Removing any point from $ (0, 1) $ disconnects it, while you can remove at most $ 1 $ point from $ (0, 1] $ without disconnecting it, and at most $ 2 $ from $ [0, 1] $. Since homeomorphism preserves connected components, the result follows.
A: Define the following topological properties, using the notion of a "cut point".
If $X$ is connected then $x$ is a cutpoint of $X$ iff $X \setminus \{x\}$ is disconnected. 
It's clear that when $f$ is a homeomorphism between $X$ and $Y$, then $x$ is a cut point for $X$ iff $f(x)$ is a cut point for $Y$.
So the number of cut points and non-cutpoints is a topological invariant. 
note that in $(0,1)$, all points are cut points,  $(0,1]$ has exactly one non-cutpoint  and $[0,1]$ has two. Moreover the last one is compact, while the first 2 are not. So the spaces are all topologically different from each other.
