Others have already shown that the general fact you state cannot be true; it would imply that for any space $X$, $X\setminus\{x\}$ and $X\setminus\{x'\}$ would always be homeomorphic.
But you have to keep in mind the structure of the proof : Suppose $f: X \to Y$ is a homeomorphism, we derive a contradiction from that. So we start with a concrete and given homeomorphism and then use the lemma:
Lemma:
Suppose $f:X \to Y$ is a homeomorphism between $X$ and $Y$. For any $a \in X$, we have that $X \setminus \{a\}$ is homeomorphic to $Y \setminus \{f(a)\}$.
Proof: just use $g : X\setminus\{a\} \to Y \setminus \{f(a)\}$ where $g$ is the restriction of $f$ (on both domain and codomain simultaneously). $g$ is continuous when $f$ is by standard properties of the subspace topology. Also $g$ is a bijection when $f$ is. And the inverse of $f$ is continuous by definition and this inverse restricted to this domain/codomain is exactly the inverse of $g$. So $g$ is also a continuous bijection with continuous inverse.
As an application:
If $f$ were a bijection between $X=(0,1)$ and $Y= [0,1)$ or $Y=[0,1]$, for some $a \in X$ we have $f(a) = 0$. Now the lemma says that $X\setminus \{a\} \simeq Y \setminus \{0\}$. Contradiction as the first space is always disconnected (regardless of $a$) and the second is connected (in both cases). So they cannot be homeomorphic, and this contradiction shows that $f$ could not exist, and $(0,1)$ is not homeomorphic to either of the other two.
Suppose $f$ is a homeomorphism between $[0,1)$ and $[0,1]$.
If $f(0) =1$, then the lemma says that $$(0,1) = [0,1)\setminus \{0\} \simeq [0,1] \setminus \{1\} = [0,1)$$ which we ruled out above. And if $f(0) = 0$ we get $$(0,1) = [0,1)\setminus \{0\} \simeq (0,1] \simeq [0,1)$$ (the latter by $h(x) = 1-x$) and again we find a situation we ruled out. So finally, $f(0) = b$ with $0 < b < 1$ but then the lemma says $$(0,1) \simeq [0,b) \cup (b,1]$$
which cannot be, by connectedness again (left is connected, right is not).
The last can be done more efficiently:
The lemma has the following corollary : (recall that a cutpoint of a connected space $X$ is a point $x \in X$ such that $X\setminus\{x\}$ is not connected, other points of $X$ are called non-cutpoints, so then $X\setminus\{x\}$ is still connected)
Corollary:
Let $f:X \to Y$ be a homeomorphism between connected spaces $X,Y$. Then if $x$ is a cutpoint of $X$, then $f(x)$ is a cutpoint of $Y$ and if $x$ is a non-cutpoint of $X$, then $f(x)$ is a non-cutpoint of $Y$.
The proof is immediate from the lemma. In particular, $X$ and $Y$ have the same number of cutpoints and non-cutpoints.
$[0,1]$ has $2$ non-cutpoints, $[0,1)$ has $1$, $(0,1)$ has none (i.e. all points are cutpoints). So no pair among them can be homeomorphic.