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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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 $\epsilon >0$, $ \exists y<x<z$ such that $\{f(y),f(z)\}\subset[0,\epsilon)$.

Without loss of generality assume that $f(y)<f(z)$.

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.
$[0,1]$ 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 .