real analysis - Non-existence of bijective, continuous function from $(0,1)$ to $[0,1]$

The problem : Give example of a continuous, onto function from $(0,1)$ to $[0,1]$. Is it possible for such a function to be one-one?

My partial solution : For the $1^{st}$ part of the question, I came up with this example -

$f : (0,1) \to [0,1]$ given by $f(x)=\sin\big({2\pi x}\big)$

This is continuous and onto, but not one-one.

What I'm asking : For the $2^{nd}$ part of the question, I feel that it should be provable that there cannot exist a continuous, onto, one-one function from $(0,1)$ to $[0,1]$ (If not, we need a counter-example). Any help regarding this proof (or what would really surprise me, a counter example)?

Thanks in advance.

Answer

Suppose $f$ is bijective, then $f$ must be strictly increasing or decreasing function. W.l.g. take $f$ to be increasing. Then $\exists x\in (0,1)$ such that $f(x)=1$. Now take $x1$, which leads to contradiction.