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
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$[0,1]$ is compact but the other two are not. Hence none of them are homeomorphic to $[0,1]$.

Now to show that $(0,1)$ and $[0,1)$ are not homeomorphic.

Suppose $f:[0,1)\to(0,1)$ is a homomorphism.

Then $f(0)\in (0,1)$ so $(0,1)\setminus \{f(0)\}$ is a disconnected set.

But $f((0,1))=(0,1)\setminus \{f(0)\}$, which is not possible as the continuous image of a connected set is connected .
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)