Continuous surjective function from [0,1] to (0,1) Is there a Continuous surjective function from [0,1] to (0,1)? 
I think there's none. Since we could take a sequence $f_n$ to approximate $1$ and by continuity and subjectivity then there's corresponding sequence $x_n$ in $[0,1]$ that approximate a certain point $a$ since $[0,1]$ is compact then that point $a$ must be in $[0,1]$.So there's $f(a)=1$ a contradiction. 
I am not sure if I am right or wrong....kind of confused by my own argument...
 A: Yes, the image of any compact set under a continuous map is itself compact. Since $[0, 1]$ is compact and $(0, 1)$ is not, there is no such function.
A: You are in the right way. The following fact will be helpful for you

Let $X$ ba a Huasdorff compact space and $f$ is continuous on $X$, then $f(x)$ is also compact.


Your proof is basically right. There is a small mistake. "You can not say that $x_n$ approximate $a$, however, $\{x_n\}$ has a subsequence which converges $a$."
A: You're on the right track. First, you can choose a sequence of points $x_n$ in $[0,1]$ so that the corresponding function values $f(x_n)$ converge to $1$; e.g., choose $x_n$ so that $f(x_n)\gt1-\frac1n$. However, it's not certain that the sequence $x_n$ converges to anything; the chosen points may jump around like crazy. However, there is a basic theorem of analysis (the Bolzano-Weierstrass Theorem) which says that any bounded sequence of real numbers has a convergent subsequence. So you can replace the original sequence $x_n$ with a subsequence converging to a point in $[0,1]$, and the corresponding values $f(x_n)$ still converge to $1$, etc.
