$(0,1), [0,1), [0,1]$ are not homeomorphic I need to show $(0,1)$, $[0,1)$ and $[0,1]$ are not homeomorphic using  intermediate value theorem (without using connectedness).
I have already did proved that $(0,1)$, $[0,1]$ are not homeomorphic but I struggle with the 2 other couples.
My proof: assume there is an homeomorphism $f:(0,1)\rightarrow[0,1]$ and take $a,b$ such that $f(a)=0, f(b)=1$. So using intermediate value theorem we can say that $f([a,b]) = [0,1]$. so $f$ isn't injective.
Will appreciate any help
Suppose $f:(0,1)\to [0,1)$ is a homeomorphism.
Then $\exists x\in(0,1)$ such that $f(x)=0$. By continuity for every  $\epsilon &gt;0$, $ \exists y&lt;x&lt;z$ such that $\{f(y),f(z)\}\subset[0,\epsilon)$.
Without loss of generality assume that $f(y)&lt;f(z)$.
Now by intermediate value property of $f$, we have $f$ on $(x,z)$ attains all the values in $(f(x),f(z))$. So there exists $z_0\in (x,z)$ such that $f(z_0)=f(y)$.
Contradicts the fact that $f$ is one-one.
Observe that if $\;f:X\to Y\;$ is a homeomorphism between two toplogical spaces and $\;x\in X\;$ then
$$\;g:X\setminus\{x\}\to Y\setminus\{f(x)\}\;,\;\;g(a)=f(a)\;\;\forall\;a\in X\setminus\{x\}\;$$
is also a homeomorphism on the respective top. spaces with the induced topology, and thus: if
$$f:[0,1)\to(0,1)\;\;\text{is a homeomorphism, then}\;\;g:(0,1)\to[(0,1)\setminus\{f(0)\}$$
is also a homeomorhpism. but this can't be since $\;(0,1)\;$ is (path) connected whereas $\;[(0,1)\setminus\{f(0)\}\;$ isn't.
Complete the other pair now (there you have path connectedness if you don't want connectedness...I don't understand why, but whatever)