Prove there is no continous bijection $f:(0,1)\to S^1$ nor $g:[0,1]\to S^1$ So I thought as follows: 
There is a continuous bijection from $[0,1)$ onto $S^1$ for example parametrization $r(x)=(\cos(2\pi x), \sin(2\pi x))$ so if bijective continuous $f:(0,1)\to S^1$ exits then there must be also some $h:(0,1)\to [0,1)$ such that $r \circ h = f$ so it is sufficient to prove that $h$ does not exists, what is easy.
For the second one $[0,1]$ is compact and $S^1$ is Hausdorff as $S^1\subset \Bbb{R}^2$ so if there exists a continuous bijection $g:[0,1]\to S^1$ then it must be homeomorphism, this cannot happened because $[0,1]$ is not homeomorphic to $S^1$
 A: Your argument for the second case is completely correct. While the first case has a bug: even if you use the word "must", the existence of $h$ does not follow immediately from the hypothesis.
A proof of the first case can be done by using the Theorem of invariance of domain which asserts that a continuous inijective map from $\mathbb R^n$ to itself is injective.
Since $S^1$ is locally homeomorphic to $\mathbb R$, then any continuous bijection $f:(0,1)\to S^1$ is open, hence $f^{-1}$ is continuous and $f$ is a homeomorphism. But $S^1$ is not homeo to $(0,1)$ (the former is compact, the later no).
The proof of the invariance of domain in dimension 1 is particularly easy: let $f:(0,1)\to \mathbb R$ continuous an injective. For any $x\in (0,1)$, we want to prove that $f(x)$ is in the interior of the image of $f$. Since $(0,1)$ is connected and $f$ is continuous then $f(0,1)$ is connected, hence an interval I of $\mathbb R$. If $f(x)$ is not an interior point then $I$ is not an open interval, and w.l.o.g. we can suppose $I=[a,b)$ and $f(x)=a$. But now we easily contradict the injectivity of $f$ near $x$ by using that $[x,x+\varepsilon)$ and $(x-\varepsilon,x]$ both maps to intervals (because $f$ is continuous) of the type $[a,something)$.
