Continuous bijection from $(0,1)$ to $[0,1]$ Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map.
No. If $f:(0,1) \to [0,1]$ were continuous and bijective, there would be a unique point $x \in (0,1)$ such that $f(x) = 1$. However, since $f$ is continuous, the intervals $[x - \varepsilon, x]$ and $[x, x + \varepsilon]$ would be mapped to intervals $[a,1]$ and $[b,1]$, say. By bijectivity we'd have $a, b \lt 1$. Thus every value strictly between $\max{\{a,b\}}$ and $1$ would be assumed at least twice, contradicting bijectivity.
There does not exist  a continuous bijection from (0,1) to [0,1]. Indeed, let $f$ be such a function. Let consider a sequence  $x_n=1-1/n$. Then from the sequence $(f(x_n))$ we can choose a subsequence $(f(x_{n_k}))$  which is convergent. Let denote this limit by $y$. Obviously, $y \in [0,1]$. Since $f^{-1}$ also is continuous, we get $f^{-1}(y)=\lim_{k \to +\infty}f^{-1}(f(x_{n_k}))=\lim_{k \to \infty}x_{n_k}=1$. But $1 \notin (0,1)$.  

Remark(Why $f^{-1}$ must be continuous under our assumption?)  By our assumption $f:(0,1)\to [0,1]$ is continuous bijection. Then  $f:(0,1)\to [0,1]$ must be injective and continuous which following invariance of domain (see, http://en.wikipedia.org/wiki/Invariance_of_domain   is  homeomorphism. Hence $f^{-1}: [0,1]\to (0,1)$ is continuous.